14.6 - Graphs of Velocity And Acceleration in SHM

In the previous section, we saw three functions:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
In this section, we will see an analysis of their graphs.

For simplicity, we will assume that $\small{\phi = 0}$. So we are going to plot the following three functions:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t)}
\\ \end{array}}$
It is shown in fig.14.29 below:

Fig.14.29

   ♦ x is plotted in red color
   ♦ v is plotted in yellow color
   ♦ a is plotted in green color
• The following assumptions are made:
   ♦ $\small{A}$ = 2.5 units
   ♦ $\small{T}$ = 6 s. So $\small{\omega = \frac{2 \pi}{T} = \frac{2 \pi}{6} = \frac{\pi}{3}}$

Based on the assumptions, we can calculate the three amplitudes:
1. Amplitude of displacement function = $\small{A}$ = 2.5 units. This is indicated by the two horizontal magenta dotted lines in the top most graph.  
2. Amplitude of velocity function = $\small{\omega A = \frac{2.5 \pi}{3}}$ = 2.62 units. This is indicated by the two horizontal magenta dotted lines in the middle graph.
3. Amplitude of acceleration function = $\small{\omega^2 A = \frac{2.5 \pi^2}{9}}$ = 2.74 units. This is indicated by the two horizontal magenta dotted lines in the bottom graph.

Let us see the connection between the graphs and the horizontal oscillating spring that we saw in fig.14.9 is section 14.1.
First we will consider displacement. It can be written in 4 steps:

1. The stop-watch is turned on at the instant when the mass is at the right extreme A. So at t=0, in red graph, abscissa is zero and ordinate is 2.5 (In our present case, We will avoid using the familiar words, x-coordinate and y-coordinate. This is because, along the horizontal axis, we are plotting t, not x. And along the vertical axis, we are plotting x, v and a, not y. See the definition for abscissa and ordinate here)

2. Let us solve the equation $\small{x(t) = A\cos(\omega t) = 0}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 1.5, 4.5, 7.5, 10.5, . . .
• These are indicated by the intersection of the red graph with the vertical red and magenta dashed lines. Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, displacement is zero. That means, those intersection points corresponds to the instants at which the mass is at the equilibrium position.

3. Let us solve the equation $\small{x(t) = A\cos(\omega t) = A}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= 2.5}$
We get: t = 0, 6, 12, . . .
• These are indicated by the intersection of the red graph with the vertical axis and the vertical white dashed lines. Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.5. That means, displacement is 2.5. That means, those intersection points corresponds to the instants at which the mass is at maximum displacement on the +ve side. The mass is at the extreme right.

4. Let us solve the equation $\small{x(t) = A\cos(\omega t) = -A}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= -2.5}$
We get: t = 3, 9, 15, . . .
• These are indicated by the intersection of the red graph with the vertical green dashed lines. Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.5. That means, displacement is −2.5. That means, those intersection points corresponds to the instants at which the mass is at maximum displacement on the −ve side. The mass is at extreme left.

---

Next we will consider velocity. It can be written in 3 steps:

1. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = 0}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 0, 3, 6, 9, 12, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ the vertical axis
   ♦ vertical green dashed lines
   ♦ vertical yellow dashed lines
Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, velocity is zero. That means, those intersection points corresponds to the instants at which the mass is stationary.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 0, the mass is at +A. Velocity here is zero
   ♦ When t = 3, the mass is at −A. Velocity here is zero
   ♦ When t = 6, the mass is at +A. Velocity here is zero
   ♦ When t = 9, the mass is at −A. Velocity here is zero
   ♦ When t = 12, the mass is at +A. Velocity here is zero
• So at extreme points, velocity is zero.

2. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = \omega A}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= \frac{2.5 \pi}{3}}$
We get: t = 4.5, 10.5, 16.5, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ vertical magenta dashed lines
Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.62. That means, velocity is 2.62. That means, those intersection points corresponds to the instants at which the mass has the maximum +ve velocity.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 4.5, the mass is at the equilibrium position, and is traveling towards the right. Velocity here is maximum. Since the velocity is towards the right, it is a +ve velocity
   ♦ When t = 10.5, the mass is at the equilibrium position, and is traveling towards the right. Velocity here is maximum. Since the velocity is towards the right, it is a +ve velocity.
• So at equilibrium position, if the mass is traveling towards the right, then the velocity will be the maximum +ve velocity.

3. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = -\omega A}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= \frac{-2.5 \pi}{3}}$
We get: t = 1.5, 7.5, 13.5, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ vertical red dashed lines
Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.62. That means, velocity is −2.62. That means, those intersection points corresponds to the instants at which the mass has the maximum −ve velocity.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 1.5, the mass is at the equilibrium position, and is traveling towards the left. Velocity here is maximum. Since the velocity is towards the left, it is a −ve velocity
   ♦ When t = 7.5, the mass is at the equilibrium position, and is traveling towards the left. Velocity here is maximum. Since the velocity is towards the left, it is a −ve velocity.
• So at equilibrium position, if the mass is traveling towards the left, then the velocity will be the maximum −ve velocity.

---

Finally, we will consider acceleration. It can be written in 3 steps:

1. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = 0}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 1.5, 4.5, 7.5, 10.5, 13.5, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical red dashed lines
   ♦ vertical magenta dashed lines
Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, acceleration is zero. That means, those intersection points corresponds to the instants at which the acceleration is zero.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 1.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum −ve
   ♦ Acceleration here is zero
When t = 4.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum +ve
   ♦ Acceleration here is zero
When t = 7.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum −ve
   ♦ Acceleration here is zero
When t = 10.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum +ve
   ♦ Acceleration here is zero
• So at equilibrium position, if the mass is traveling towards the right, then the velocity will be the maximum +ve velocity. And acceleration will be zero.

2. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = \omega^2 A}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= \frac{2.5^2 \pi}{9}}$
We get: t = 3, 9, 15, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical green dashed lines
• Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.74. That means, acceleration is 2.74. That means, those intersection points corresponds to the instants at which the mass has the maximum +ve acceleration.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 3,
   ♦ The mass is at −A
   ♦ Velocity here is zero
   ♦ Acceleration here is +ve maximum because, it is towards right
When t = 9,
   ♦ The mass is at −A
   ♦ Velocity here is zero
   ♦ Acceleration here is +ve maximum because, it is towards right
• So at the −ve extreme point, velocity is zero and acceleration is +ve maximum

3. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = -\omega^2 A}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= \frac{(-1)2.5^2 \pi}{9}}$
We get: t = 0, 6, 12, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical axis
   ♦ vertical white dashed lines
• Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.74. That means, acceleration is −2.74. That means, those intersection points corresponds to the instants at which the mass has the maximum −ve acceleration.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 0,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
When t = 6,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
When t = 12,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
• So at the +ve extreme point, velocity is zero and acceleration is −ve maximum

---

In general, we can write 3 points:
1. When the mass is at equilibrium position,
   ♦ velocity (+ve or −ve) is maximum
   ♦ acceleration is zero
This is indicated by the red and magenta vertical dashed lines
2. When the mass is at +A,
   ♦ velocity is  zero
   ♦ acceleration is maximum −ve
This is indicated by the vertical axis and white vertical dashed lines
3. When the mass is at −A,
   ♦ velocity is  zero
   ♦ acceleration is maximum +ve
This is indicated by the green vertical dashed lines

---

Now we will see a solved example

Solved example 14.5
A body oscillates with SHM according to the equation (in SI units) $\small{x = 5 \cos\left(2\pi t + \frac{\pi}{4} \right)}$.
At t = 1.5 s, calculate the (a) displacement (b) speed and (c) acceleration of the body.
Solution:
1. For any SHM, we can write three equations:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$

2. In our present case, we are given the first one as:
$\small{x = 5 \cos\left(2\pi t + \frac{\pi}{4} \right)}$

3. Comparing them, we can write:
A = 5 m, $\small{\omega = 2\pi}$ rad/s and $\small{\phi = \frac{\pi}{4}}$ rad

4. So for this problem, the three equations can be written as:
(a) Displacement:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {=}    &{ 5{\rm{(m)}}\cos\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{5{\rm{(m)}}\cos\left[2\pi t(\rm{rad}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {=}    &{5{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$
(b) velocity:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(t)}    & {=}    &{-2\pi(\rm{rad/s})5{\rm{(m)}}\sin\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

(c) acceleration:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a(t)}    & {=}    &{-[2\pi(\rm{rad/s})]^2~5{\rm{(m)}}\cos\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

4. So for this problem, the three equations are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

5. Now we can write the required quantities
(a) Displacement:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(1.5)}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[2\pi (1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{5(m)\cos\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-3.535{\rm{(m)}}}
\\ \end{array}}$

• Recall that cosine of any angle is a mere number. It does not have any unit.

(b) velocity:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(1.5)}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi(1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{22.214{\rm{(m/s)}}}
\\ \end{array}}$

• Recall that, sine of any angle is a mere number. It does not have any unit.

• Also recall that, radians is the ratio of two lengths. So it is a dimensionless quantity. Therefore, we do not write it as a unit in the final result. An angle measured in radians, is a mere number. Details here.

(c) acceleration:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a(1.5)}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi (1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{139.58{\rm{(m/s^2)}}}
\\ \end{array}}$

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In the next section, we will see the force law for SHM.

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14.5 - Velocity and Acceleration in Simple Harmonic Motion

In the previous section, we saw simple harmonic motion (S.H.M). In this section, we will see velocity and acceleration in S.H.M.

Velocity in simple harmonic motion

• In fig.14.25 below, the magenta sphere is in uniform circular motion.
   ♦ Radius of the circle is A
   ♦ Angular speed of the sphere is $\small{\omega}$

Fig.14.25

• We know that, the "projection of the sphere on the x-axis" will be in SHM between (−A,0) and (A,0). We want an expression which can be used to calculate the "velocity of this projection" at any instant t.
This can be done in 4 steps:

1. Let at any instant t, the position of the sphere be P.
We know that, at any instant, the tangential velocity of the sphere will be $\small{\omega\,A}$. This is indicated by the white tangential arrow at P, in the fig.14.25 above.

2. Now we drop two perpendicular green dashed lines, onto the x-axis.
• One through P and the other through the head of $\small{\omega\,A}$
• These two perpendicular lines will give the projection of $\small{\omega\,A}$ on the x-axis.

3. This projection is the velocity of the SHM at the instant t. It is denoted as $\small{v(t)}$. So our next aim is to find the expression for $\small{v(t)}$

4. Fig.14.26 below shows an enlarged view of the first quadrant.

Fig.14.26

• The white dotted line is parallel to the x-axis and it passes through P. So by the rule of alternate angles, angle between OP and the dotted line is $\small{(\omega t~+~\phi)}$
• The tangential velocity $\small{\omega\,A}$ is perpendicular to OP. So the angle between the tangential velocity and the dotted line will be $\small{\left[\frac{\pi}{2} - (\omega t + \phi) \right]}$
• So the horizontal component of the tangential velocity will be $\small{\omega\,A \cos\left[\frac{\pi}{2} - (\omega t + \phi) \right]\,=\,\omega\,A \sin (\omega t + \phi)}$
See identity 5 in the list of trigonometric identities.
• But in the final expression, we must include a −ve sign because, the horizontal component is acting towards the −ve side of the x-axis.
• Thus we get: $\small{v(t)\,=\,-\omega\,A \sin (\omega t + \phi)}$

---

Note:
In math classes, it is proved that, derivative of displacement, w.r.t time, will give velocity. This is true in the case of SHM also.

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{\frac{dx}{dt}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{v(t)}    & {~=~}    &{\frac{d\left[A\cos(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ \end{array}}$

• This is the same expression that we obtained above

---

Acceleration in simple harmonic motion

• In fig.14.27 below, we consider the same magenta sphere again.
   ♦ Radius of the circle is A
   ♦ Angular speed of the sphere is $\small{\omega}$

Fig.14.27

• We know that, the "projection of the sphere on the x-axis" will be in SHM between (−A,0) and (A,0). We want an expression which can be used to calculate the "acceleration of this projection" at any instant t.
This can be done in steps:

1. Let at any instant t, the position of the sphere be P.
We know that, at any instant, the centripetal acceleration experienced by the sphere will be $\small{v^2\,A}$ or $\small{\omega^2\,A}$. This is indicated by the white radial arrow at P, in the fig.14.27 above.

2. Now we drop two perpendicular green dashed lines, onto the x-axis.
• One through P and the other through the head of $\small{\omega\,A}$
• These two perpendicular lines will give the projection of $\small{\omega^2\,A}$ on the x-axis.

3. This projection is the acceleration of the SHM at the instant t. It is denoted as $\small{a(t)}$. So our next aim is to find the expression for $\small{a(t)}$

4. Fig.14.28 below shows an enlarged view of the first quadrant.

Fig.14.28

• The white dotted line is parallel to the x-axis and it passes through the head of $\small{\omega^2\,A}$. So the angle between $\small{\omega^2\,A}$ and the dotted line is $\small{(\omega t~+~\phi)}$
• So the horizontal component of $\small{\omega^2\,A}$ will be $\small{\omega^2\,A \cos(\omega t + \phi)}$
• But in the final expression, we must include a −ve sign because, the horizontal component is acting towards the −ve side of the x-axis.
• Thus we get: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$

---

Note:
In math classes, it is proved that, derivative of velocity, w.r.t time, will give acceleration. This is true in the case of SHM also.

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{a(t)}    & {~=~}    &{\frac{dv}{dt}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{a(t)}    & {~=~}    &{\frac{d\left[-A \omega\sin(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$

• This is the same expression that we obtained above

---

Now we will see two important properties related to acceleration.
Property 1:
This can be explained in 4 steps:
1. We obtained: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$
2. But in the previous section, we saw that: $\small{x(t)\,=\,A \cos (\omega t + \phi)}$
3. So we can write: $\small{a(t)\,=\,-\omega^2\,x(t)}$
4. Suppose that, $\small{\omega}$ is a constant. Then we can write:
acceleration is proportional to displacement.

Property 2:
This can be explained in steps:
1. In property 1, we saw that:
Acceleration = A constant  × displacement
2. But there is a −ve sign in the expression for acceleration. So we can write:
   ♦ If the displacement is −ve, acceleration will be +ve.
   ♦ If the displacement is +ve, acceleration will be −ve.

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Let us see how the two properties are applicable to the horizontal oscillating spring that we saw in fig.14.9 is section 14.1. It can be written as four cases.

Case 1
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=0 to x=A
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
• That is, acceleration will be acting towards the −ve side of the x-axis
• That means, acceleration will be acting opposite to the direction of motion.
• That means, the mass will be experiencing deceleration.
• Indeed, due to this deceleration, the mass comes to a stop at x = A
3. Also, the deceleration experienced by the mass is not uniform.
• Greater the distance from the equilibrium position, greater will be the deceleration. That means, as the mass approaches x=A, it will be experiencing more and more deceleration.

Case 2
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=A to x=0
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
• That is, acceleration will be acting towards the −ve side of the x-axis
• That means, acceleration will be acting in the direction of motion.
• Indeed, due to this acceleration, the mass will have the maximum possible velocity at x = 0
3. Also, the acceleration experienced by the mass is not uniform.
• Lesser the distance from the equilibrium position, lesser will be the acceleration. That means, as the mass approaches x = 0, it will be experiencing less and less acceleration.

Case 3
This is the case when the mass is traveling from x=0 to x=−A
The reader may write all the 3 steps in detail

Case 4
This is the case when the mass is traveling from x=−A to x=0
The reader may write all the 3 steps in detail

---

We have seen three functions:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
In the next section, we will see an analysis of their graphs

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14.4 - Simple Harmonic Motion

In the previous section, we saw that displacement from the equilibrium position can be specified using sine or cosine functions. In this section, we will see simple harmonic motion.

1. Consider a particle subjected to oscillation along the x-axis. Suppose that, the oscillation satisfies the following conditions:
(i) -`A and A are the extreme points.
(ii) The origin O is the equilibrium point.
(iii) Displacement from the origin can be specified using a sinusoidal function.
(Sine and cosine functions are examples for sinusoidal functions)
2. We have seen that the particle attached to a horizontal spring can satisfy the above three conditions. (see section 14.1)
Let us observe this oscillation. Assume that, the stop-watch is turned on when the particle is at O
3. Then, in addition to those three conditions, the particle satisfies more conditions:
(iv) When $\small{t~=~0}$, the particle is at O, and the velocity is maximum.
(v) When $\small{t~=~\frac{T}{4}}$, the particle is at +A, and the velocity is zero.
(vi) When $\small{t~=~\frac{T}{2}}$, the particle is at O, and the velocity is maximum.
(vii) When $\small{t~=~\frac{3T}{4}}$, the particle is at -`A, and the velocity is zero.
(viii) When $\small{t~=~T}$, the particle is at O, and the velocity is maximum.
(ix) When $\small{t~=~\frac{5T}{4}}$, the particle is at +A, and the velocity is zero.
so on . . .
4. If the nine conditions are satisfied, then that oscillation is called a simple harmonic motion. It is abbreviated as SHM.
5. In the previous sections, we became familiar with the function:
$\small{x(t)~=~A \cos\left(\omega t \,+\,\phi \right)}$
    ♦ $\small{x(t)}$ is the displacement $\small{x}$ as a function of time $\small{t}$
    ♦ $\small{A}$ is the amplitude
    ♦ $\small{\omega}$ is the angular frequency
    ♦ $\small{\omega t \,+\,\phi}$ is the phase. It depends on time.
    ♦ $\small{\phi}$ is the phase constant
• This is a sinusoidal function which can represent a SHM.

---

Let us see some examples:

Example 1:
In fig.14.21 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.21

From the graph, we get two information:
(i) Information about period T
• Both red and green reach maximums at the same instants.
• Both red and green reach minimums at the same instants
• Both red and green reach zero at the same instants
• So red and green have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
Green: $\small{x(t)~=~3 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The instants corresponding to maximums of red, can be obtained by solving the equation:
$\small{4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$ 

The instants corresponding to minimums of red, can be obtained by solving the equation:
$\small{-4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The instants corresponding to zeros of red, can be obtained by solving the equation:
$\small{0~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The same procedure can be used for green.

Always remember the relation between T and $\small{\omega}$, which is: $\small{\omega~=~\frac{2 \pi}{T}}$

So for the red, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\omega t}    & {~=~}    &{\frac{\pi t}{6}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\omega}    & {~=~}    &{\frac{\pi}{6}~=~\frac{2 \pi}{T}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{T}    & {~=~}    &{12~\text{seconds}}    \\
\end{array}}$

In the same way, we can find the period of green also.

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of red is 4 units.  
• The two magenta horizontal dashed lines indicate that, the amplitude of green is 3 units.

◼ Based on this example, we can write:
Two simple harmonic motions may have the same phase $\small{\left(\omega t \,+\,\phi \right)}$. But they can have different amplitudes.

Example 2:
In fig.14.22 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.22

From the graph, we get two information:
(i) Information about period T
• Red and green do not reach maximums at the same instants.
• Red and green do not reach minimums at the same instants
• Red and green do not reach zero at the same instants
• So red and green do not have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6}  \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,-\,\frac{\pi}{4} \right)}$

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.  

◼ Based on this example, we can write:
Two simple harmonic motions may have the same amplitude $\small{A}$. But they can have different  phases.

Example 3:
In fig.14.23 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.23

From the graph, we get two information:
(i) Information about period T
• Consider the magenta vertical dashed line. Both red and green reach maximum at the time indicated by this dashed line.
• The "horizontal distance between y-axis and the magenta vertical dashed line" indicates a time duration. Within this time duration, green completes two cycles. Red completes only one cycle.
• So it is clear that, $\small{T_{\text{green}}}$ is half of $\small{T_{\text{red}}}$
• So red and green do not have the same angular frequency $\small{\left(\omega \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{4}  \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{2} \,-\,\frac{\pi}{4} \right)}$

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.  

◼ Based on this example, we can write:
Two simple harmonic motions may have the same amplitude $\small{A}$. But they can have different angular frequencies.

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Now we will see a solved example:

Solved example 14.4
Which of the following function of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic ? Give the period for each case
(i) $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
(ii) $\small{\sin^2 \left(\omega t \right)}$
Solution:
Part (i):
1. We are given: $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}    & {~=~}    &{\sin\left(\omega t \right)~-~\sin\left(\frac{\pi}{2} - \omega t \right)}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2 \cos\left(\frac{\pi}{4} \right)\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2 \left(\frac{1}{\sqrt 2} \right)\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
\end{array}}$

◼ Remarks:

• 1 (magenta color):
Here we use the identity 6 in the list of trigonometric identities.
• 2 (magenta color):
Here we use the identity 20(d) in the list of trigonometric identities.

2. Based on this standard form, we can write:
The given expression represents a SHM with,
    ♦ Amplitude $\small{\sqrt 2}$ units
    ♦ Angular frequency $\small{\omega}$
    ♦ Phase constant $\small{- \frac{\pi}{4}}$

3. The standard form obtained in (1) can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left[2\pi~+~\left(\omega t - \frac{\pi}{4} \right) \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t + \frac{7\pi}{4} \right)}    \\
\end{array}}$

• So for the given SHM, the phase constant $\small{- \frac{\pi}{4}}$ is same as $\small{\frac{7\pi}{4}}$.

Part (ii):
1. We are given: $\small{\sin^2 \left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin^2 \left(\omega t \right)}    & {~=~}    &{\frac{1\,-\,\cos\left(2\omega t \right)}{2}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{1}{2}~-~\frac{\cos\left(2\omega t \right)}{2}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{1}{2}~-~\frac{1}{2}\,\cos\left(2\omega t \right)}    \\
\end{array}}$

◼ Remarks:

• 1 (magenta color):
Here we use the identity 14 in the list of trigonometric identities.

2. In fig.14.24 below, two graphs are shown.

Fig.14.24

• Green represents: $\small{x(t)~=~0.5 \cos\left( 2\omega t \right)}$
    ♦ The two horizontal magenta dashed lines indicate that, the amplitude of green is 0.5 units.
    ♦ This graph is familiar to us. The equilibrium point is at O.

• Red represents: $\small{x(t)~=~0.5\,-\,0.5 \cos\left( 2\omega t \right)}$
    ♦ This is also a sinusoidal curve.
    ♦ The horizontal yellow dashed line and the x-axis indicate that, the distance between extreme points is 1 unit.
    ♦ So the amplitude of red is 1/2, which is 0.5 units.
    ♦ The upper horizontal magenta dashed line divides the red into two equal upper and lower parts. This dashed line passes through 0.5 on the y-axis. That means, equilibrium point of red is 0.5 units away from the origin.

3. Comparing the green with the standard form, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2\omega}    & {~=~}    &{\frac{2\pi}{T}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\omega}    & {~=~}    &{\frac{\pi}{T}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{T}    & {~=~}    &{\frac{\pi}{\omega}}    \\
\end{array}}$

---

In the next section, we will see velocity and acceleration in simple harmonic motion.

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14.3 - Displacement in Simple Pendulum

In the previous section, we saw the displacement in the case of a spring oscillating vertically. In this section, we will see the displacement in a simple pendulum.

Let us first see the details of the oscillation. It can be written in 6 steps:
1. In fig.14.15 (a) below, a magenta mass m, is attached to a string. The other end of the string is fixed to a rigid ceiling. We assume that, the resistance due to the surrounding air is negligible.
• Mass m should not be very heavy. It need not be much heavier than that required to keep the string taut.
• Length L is measured from the point of support O, to the center of the mass m.
• The mass is said to be at the equilibrium position in this fig.a

Fig.14.15

2. In fig.14.15 (b), the mass is pulled to the right by a horizontal distance A. From this position, the mass is released from rest. It will then swing towards left. Even after traveling a horizontal distance 'A', it will continue to swing towards the left. That is., even after reaching the equilibrium position, it will continue to swing towards the left. This is shown in fig.c.

3. But once the equilibrium position is passed, the mass will begin to experience a resistive force. The mass is able to overcome the resistive force, and travel a horizontal distance A. Once it reaches this point, it stops. That is., it's velocity becomes zero. This is shown in fig.d.

4. Then it starts the reverse journey. In the reverse journey, even after passing the equilibrium position, it will continue to swing towards the right. This is shown in fig.e.

5. But once the equilibrium position is passed, the mass will begin to experience a resistive force. The mass is able to overcome the resistive force, and reach up to the initial point. Once it reaches the initial point, it stops. That is., it's velocity becomes zero. This is shown in fig.f.

6. At this point, one cycle is complete. Then it again starts the swing towards the left. This process continues giving rise to continuous oscillation.

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Displacement

The above 6 steps give us a basic understanding about oscillation of the simple pendulum. Now we will derive an expression for displacement. It can be done in 4 steps:
1. The mass will be swinging along an arc of radius L. This is shown in fig.14.16 below.

Fig.14.16

2. The stop-watch is turned on, at the instant the mass is released from the right extreme point.
• At any instant, when the reading in the stop-watch is 't', let the mass be at M.
• At that instant, let the angle which OM makes with the vertical be $\small{\theta}$

3. Drop the perpendicular MN from M onto the vertical.
• Now we have a right triangle OMN.
• In this right triangle,
$\small{\sin\theta~=~\frac{MN}{L}~\Rightarrow MN = L \sin \theta}$
• So we get a method to write the horizontal displacement of the mass. We can write:
$\small{x(t)~=~L\,\sin\theta}$

4. Here we do not need to consider the phase constant because, at any instant, the angle $\small{\theta}$ is measured from the vertical

---

We have seen three types of oscillations. Horizontal spring, vertical spring and simple pendulum. Based on those discussions, we can say that, the displacement from equilibrium position can be specified using sine function or cosine function.

Let us see a solved example:

Solved example 14.2
Fig.14.17 below, depicts two circular motions. The radius of the circle, period of revolution, initial position and the direction of revolution are indicated in the figures. Obtain the expression for horizontal displacement of the rotating particle P in each case.

Fig.14.17

Solution:
Part (i):
1. Consider fig.14.18(a) below:

Fig.14.18

• At the instant when the stop-watch is turned on, the particle is at M(t=0)
• Here, OM(t=0) is already making an angle of 45 deg ($\small{\frac{\pi}{4}}$  radians) with the +ve side of the x-axis.

2. At the instant when the reading in the stop-watch is ‘t’, the particle is at M.
• Here, OM makes an angle $\small{\theta}$ with the initial position.
• Drop the perpendicular MN, from M onto the x-axis.

3. So in the right triangle OMN, the angle MON = $\small{\left(\frac{\pi}{4}~+~\theta \right)}$
• Therefore, the horizontal displacement ON
= $\small{OM\,\cos\left(\frac{\pi}{4}~+~\theta \right)~=~A\,\cos\left(\frac{\pi}{4}~+~\theta \right)}$

4. We know that $\small{\theta~=~\frac{2\pi t}{T}}$
• So the horizontal displacement can be written as:
$\small{A\,\cos\left(\frac{\pi}{4}~+~\frac{2\pi t}{T} \right)}$
• From fig.14.17(a), we have: T = 4 s. So we can write the horizontal displacement in standard form:
$\small{x(t)~=~A\,\cos\left(\frac{\pi}{4}~+~\frac{2\pi t}{4} \right)}$
$\small{\Rightarrow x(t)~=~A\,\cos\left(\frac{2\pi t}{4}~+~\frac{\pi}{4} \right)}$

5. Based on this standard form, we can write:
• The oscillation in fig.14.17(a) has
    ♦ Amplitude A
    ♦ Period 4 s
    ♦ Phase constant $\small{\frac{\pi}{4}}$  

Part (ii):
1. Consider fig.14.18(b) above.
• At the instant when the stop-watch is turned on, the particle is at M(t=0)
• Here, OM(t=0) is already making an angle of $\small{\frac{-3\pi}{2}}$  radians with the +ve side of the x-axis. Note that in this case, revolution is in the clockwise direction. So we measure the angle also in the clockwise direction, from the +ve side of the x-axis. Consequently, the angle is −ve.

2. At the instant when the reading in the stop-watch is ‘t’, the particle is at M.
• Here, OM makes an angle $\small{\theta}$ with the initial position.
• So the total angle measured from the +ve side of the x-axis
= $\small{\left[\frac{-3\pi}{2} - \theta \right]~=~\left[-\left(\frac{3\pi}{2} + \theta \right) \right]}$

3. Drop the perpendicular MN, from M onto the x-axis.
• Now we can write:
x-coordinate of N = x-coordinate of M
= OM × cosine of $\small{\left[-\left(\frac{3\pi}{2} + \theta \right) \right]}$
• Therefore, the horizontal displacement ON
= $\small{OM\,\cos\left[-\left(\frac{3\pi}{2} + \theta \right) \right]~=~B\,\cos\left[\frac{3\pi}{2} + \theta  \right]}$

• Cosine of a −ve angle is +ve. See identity 2 in the list of trigonometric identities.

$\small{\Rightarrow B\,\cos\left[2 \pi~-~\frac{\pi}{2} + \theta  \right]~=~B\,\cos\left[2 \pi~+~\left(\theta - \frac{\pi}{2} \right) \right]~=~B\,\cos\left(\theta - \frac{\pi}{2} \right)}$

4. We know that $\small{\theta~=~\frac{2\pi t}{T}}$
• So the horizontal displacement can be written as:
$\small{B\,\cos\left(\frac{2\pi t}{T} - \frac{\pi}{2} \right)}$
• From fig.14.17(b), we have: T = 30 s. So we can write the horizontal displacement in standard form:
$\small{x(t)~=~B\,\cos\left(\frac{2\pi t}{30}~-~ \frac{\pi}{2} \right)}$

5. Based on this standard form, we can write:
• The oscillation in fig.14.17(b) has
    ♦ Amplitude B
    ♦ Period 30 s
    ♦ Phase constant $\small{\frac{-\pi}{2}}$

Solved example 14.3
Fig.14.19 below, depicts two circular motions. The radius of the circle, period of revolution, initial position and the direction of revolution are indicated in the figures. Obtain the expression for horizontal displacement of the rotating particle P in each case.

Fig.14.19

Solution:
Part (i):
1. Consider fig.14.20(a) below:

Fig.14.20

• At the instant when the stop-watch is turned on, the particle is at M(t=0)
• Here, OM(t=0) is already making an angle of $\small{\frac{-\pi}{2}}$  radians with the +ve side of the x-axis. Note that in this case, revolution is in the clockwise direction. So we measure the angle also in the clockwise direction, from the +ve side of the x-axis. Consequently, the angle is −ve.

2. At the instant when the reading in the stop-watch is ‘t’, the particle is at M.
• Here, OM makes an angle $\small{\theta}$ with the initial position.
• So the total angle measured from the +ve side of the x-axis
= $\small{\left[\frac{-\pi}{2} - \theta \right]~=~\left[-\left(\frac{\pi}{2} + \theta \right) \right]}$

3. Drop the perpendicular MN, from M onto the x-axis.
• Now we can write:
x-coordinate of N = x-coordinate of M
= OM × cosine of $\small{\left[-\left(\frac{\pi}{2} + \theta \right) \right]}$
• Therefore, the horizontal displacement ON
= $\small{OM\,\cos\left[-\left(\frac{\pi}{2} + \theta \right) \right]~=~3\,\cos\left[\frac{\pi}{2} + \theta  \right]}$

• Cosine of a −ve angle is +ve. See identity 2 in the list of trigonometric identities.

4. We know that $\small{\theta~=~\frac{2\pi t}{T}}$
• So the horizontal displacement can be written as:
$\small{3\,\cos\left(\frac{2\pi t}{T} + \frac{\pi}{2} \right)}$
• From fig.14.19(a), we have: T = 2 s. So we can write the horizontal displacement in standard form:
$\small{x(t)~=~3\,\cos\left(\frac{2\pi t}{2}~+~ \frac{\pi}{2} \right)}$

5. Based on this standard form, we can write:
• The oscillation in fig.14.19(a) has
    ♦ Amplitude 3
    ♦ Period 2 s
    ♦ Phase constant $\small{\frac{\pi}{2}}$

6. We obtained:
$\small{x(t)~=~3\,\cos\left(\frac{2\pi t}{2}~+~ \frac{\pi}{2} \right)}$
$\small{\Rightarrow x(t)~=~3\,\cos\left(\pi t~+~ \frac{\pi}{2} \right)}$
• This is same as:
$\small{x(t)~=~-3\,\sin\left(\pi t \right)}$
See identity 9(a) in the list of trigonometric identities.

Part (ii):
1. Consider fig.14.20(b) above.
• At the instant when the stop-watch is turned on, the particle is at M(t=0)
• Here, OM(t=0) is already making an angle of 180 deg ($\small{\pi}$  radians) with the +ve side of the x-axis.

2. At the instant when the reading in the stop-watch is ‘t’, the particle is at M.
• Here, OM makes an angle $\small{\theta}$ with the initial position.
• Drop the perpendicular MN, from M onto the x-axis.

3. Now we can write:
x-coordinate of N = x-coordinate of M
= OM × cosine of $\small{\left(\pi + \theta \right)}$
• Therefore, the horizontal displacement ON
= $\small{OM\,\cos\left(\pi + \theta \right)}$

4. We know that $\small{\theta~=~\frac{2\pi t}{T}}$
• So the horizontal displacement can be written as:
$\small{2\,\cos\left(\pi~+~\frac{2\pi t}{T} \right)}$
• From fig.14.19(a), we have: T = 4 s. So we can write the horizontal displacement in standard form:
$\small{x(t)~=~2\,\cos\left(\pi~+~\frac{2\pi t}{4} \right)}$
$\small{\Rightarrow x(t)~=~2\,\cos\left(\frac{2\pi t}{4}~+~\pi \right)}$

5. Based on this standard form, we can write:
• The oscillation in fig.14.19(a) has
    ♦ Amplitude 2 m
    ♦ Period 4 s
    ♦ Phase constant $\small{\pi}$

6. We obtained:
$\small{x(t)~=~2\,\cos\left(\frac{2\pi t}{4}~+~ \pi \right)}$
$\small{\Rightarrow x(t)~=~2\,\cos\left(\pi~+~\frac{\pi t}{2}  \right)}$
• This is same as:
$\small{x(t)~=~-2\,\cos\left(\frac{\pi t}{2} \right)}$
See identity 9(e) in the list of trigonometric identities.

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In the next section, we will see simple harmonic motion.

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14.2 - Displacement in Vertical Spring

In the previous section, we saw the displacement in the case of a spring oscillating horizontally. In this section, we will see the displacement of a spring oscillating vertically.

Let us first see the details of the oscillation. It can be written in 6 steps:
1. In fig.14.9 (a) below, a red block of mass m, is attached to a spring. The other end of the spring is fixed to a rigid ceiling. We assume that, the resistance due to the surrounding air is negligible.

Fig.14.9

2. In fig.14.9 (b), the block is pulled downwards by a distance A. From this position, the block is released from rest. It will then travel upwards. Even after traveling a distance 'A', it will continue to travel upwards. That is., even after passing the point y= 0, it will continue to travel upwards. This is shown in fig.c.

3. But once the point y = 0 is passed, the block will begin to experience a resistive force (in addition to gravity). The block is able to overcome the resistive force, and reach up to y = +A. Once it reaches y=A, it stops. That is., it's velocity becomes zero. This is shown in fig.d.

4. Then it starts the reverse journey. In the reverse journey, even after passing the point y=0, it will continue to travel downwards. This is shown in fig.e.

5. But once the point y = 0 is passed, the block will begin to experience a resistive force. The block is able to overcome the resistive force, and reach up to y = −A. Once it reaches y=−A, it stops. That is., it's velocity becomes zero. This is shown in fig.f.

6. At this point, one cycle is complete. Then it again starts the upward journey. This process continues giving rise to continuous oscillation.

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Displacement

The above 6 steps give us a basic understanding about oscillation. Now we will derive an expression for displacement. It can be done in 9 steps:
1. We saw that, the red block is oscillating between y=−A and y=+A. Let us mark those two points as P and Q respectively, on the y-axis. This is shown in fig.14.10 below:

Fig.14.10

• The coordinates of P and Q are (0,−A) and (0,A) respectively. So O is the equilibrium position of the block.

2. Draw the red circle with center at O and radius equal to A. The magenta sphere is performing uniform circular motion along the red circle.
• The angular velocity of the magenta sphere is $\small{\omega}$.
• That means, the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.

3. As we did in the previous section, let us try to write a general method to find the displacement.

(i) In fig.14.11(a) below, M is in the IV quadrant.
    ♦ On OM, Mark M' such that, OM' = 1 unit.
    ♦ Drop the perpendicular M'N' onto the y-axis.

Fig.14.11

• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM'}{OM}~=~\frac{ON'}{ON}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the y-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the y-coordinate of M' is $\small{\sin \theta}$ (Details here)
• So we get:
$\small{ON~=~A \left(\frac{\sin \theta}{1} \right)~=~A\,\sin \theta}$

(ii)  In fig.14.11(b) above, M is in the I quadrant.
    ♦ On OM, Mark M' such that, OM' = 1 unit.
    ♦ Drop the perpendicular M'N' onto the y-axis.
• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM'}{OM}~=~\frac{ON'}{ON}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the y-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the y-coordinate of M' is $\small{\sin \theta}$
• So we get:
$\small{ON~=~A \left(\frac{\sin \theta}{1} \right)~=~A\,\sin \theta}$

(iii) In fig.14.11(c) above, M is in the II quadrant.
We can write similar steps and obtain the same result.

(iv) In fig.14.11(d) above, M is in the III quadrant.
We can write similar steps and obtain the same result.
• So whichever be the quadrant, the vertical displacement of the magenta sphere will be $\small{A\,\sin \theta}$

• Let us see an example:
Suppose that $\small{\theta~=~560 \deg}$
Then the red sphere is in the III quadrant. We get:
Vertical displacement = $\small{ON~=~A\,\sin(560)~=~A(-0.3420)}$
Indeed, the vertical displacement will be −ve in the III quadrant because, N lies on the −ve side of the y-axis

4. We see that, this method is very effective to write the vertical displacement of the magenta sphere. But we want the vertical displacement of the red block.
• So we assume that:
   ♦ At the instant when the block is released from P, the sphere starts the revolution from P
   ♦ At the instant when the block reaches O, the sphere reaches R
   ♦ At the instant when the block reaches Q, the sphere also reaches Q
   ♦ At the instant when the block returns through O, the sphere reaches S
   ♦ At the instant when the block returns back at P, the sphere also reaches back at P

5. That means, the time period (T) for one oscillation of the block is same as the time for one revolution of the sphere. Using this information, we can write an expression for $\small{\theta}$
• Time for one revolution of the sphere = T seconds
⇒ Time for $\small{2 \pi}$ radians = T seconds
⇒ Time for 1 radian = $\small{\frac{T}{2 \pi}}$ seconds
⇒ Angular distance covered in 1 second  = $\small{\frac{2 \pi}{T}}$ radian
• The stop-watch is turned on at the instant when the block is released from P. At that same instant, the sphere starts the revolution from the same point P.
• Consider the instant at which the reading in the stop-watch is t. Let at that instant, the sphere be at M.
• So when the reading is t, the angular distance covered is $\small{\theta}$
• Angular distance covered in 1 second  = $\small{\frac{2 \pi}{T}}$ radian
⇒ Angular distance covered in t seconds  = $\small{\frac{2 \pi t}{T}}$ radian
⇒ $\small{\theta}$  = $\small{\frac{2 \pi t}{T}}$ radian
• So we can write:
At any time t,
The vertical displacement of the sphere from O
= The vertical displacement of the block from O
= $\small{A\,\sin\theta~=~A\,\sin\left(\frac{2 \pi t}{T} \right)}$
• That means:
$\small{x(t)~=~A\,\sin\left(\frac{2 \pi t}{T} \right)}$

6. We derived the equation: $\small{x(t)~=~A\,\sin\theta~=~A\,\sin\left(\frac{2 \pi t}{T} \right)}$.
• If we want, we can use $\small{\omega}$ instead of $\small{T}$.
• We have:
the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.
• So in $\small{t}$ seconds, the sphere will subtend $\small{\omega\,t}$ radians at O
• But the angle subtended by the sphere, at O, in $\small{t}$ seconds, is $\small{\theta}$.
• That means: $\small{\theta~=~\omega\,t}$
• Thus we get:
$\small{x(t)~=~A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)~=~A\,\cos\left(\omega\,t \right)}$.

7. Here, the only variable on the R.H.S is $\small{t}$.
   ♦ $\small{t}$ is the independent variable
   ♦ $\small{x}$ is the dependent variable
• So while drawing the graph, we must plot $\small{t}$ along the x-axis and $\small{x}$ along the y-axis.

8. One such graph is shown in fig.14.12 below.
   ♦ A is assumed to be 3 units
   ♦ T is assumed to be 10 s

Fig.14.12

• We get:
$\small{x(t)~=~A\,\sin\left(\frac{2 \pi t}{T} \right)~=~(3)\,\sin\left(\frac{2 \pi t}{10} \right)~=~(3)\,\sin\left(\frac{\pi t}{5} \right)}$

• Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~3~=~(3)\,\sin\left(\frac{\pi t}{5} \right)}$
$\small{\Rightarrow 1~=~\sin\left(\frac{\pi t}{5} \right)}$

• This is a trigonometrical equation. We have seen the method to solve such equations, in our math classes (Details here).
• We get:
t = 2.5, 12.5, 22.5, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 3

(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-3~=~(3)\,\sin\left(\frac{\pi t}{5} \right)}$
$\small{\Rightarrow -1~=~\sin\left(\frac{\pi t}{5} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 7.5, 17.5, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −3

(iii) To find the instants at which the red block reaches the equilibrium point, we need to solve the equation:
$\small{x(t)~=~0~=~(3)\,\sin\left(\frac{\pi t}{5} \right)}$
$\small{\Rightarrow 0~=~\sin\left(\frac{\pi t}{5} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 0, 5, 10, 15, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.

(iv) We can obtain time period T from the graph:
• (2.5,3) and (12.5,3) are two consecutive +ve extreme points.
• The difference between their x-coordinates is (12.5 − 2.5) = 10 s
• This 10s is the time period T that we assumed to draw the graph.
• The red block can be assumed to start from the +ve extreme at t = 2.5 s. It then travels to the −ve extreme and returns to the +ve extreme at t = 12.5 s. That means, the time for one complete cycle is 10 s 

---

Phase constant

This can be explained in 3 steps:
1. Consider fig.14.10 again. We turned the stop-watch on, when the red sphere was at P. Then we noted the time 't' at which the red sphere is at the arbitrary point M. The angle MOP at time 't' is denoted as '$\small{\theta}$'

2. The same fig.14.10 is modified and shown again in fig.14.13(a) below:

Fig.14.13

• Here, the stop-watch is turned on, when the sphere is at B. At B, the line OB already makes an angle $\small{\phi}$ with the x-axis.
• So at time 't', the sphere is at the arbitrary point M and the line OM makes an angle of $\small{\theta + \phi}$ with the x-axis.
• Now consider fig.14.13(b) above.
    ♦ On OM, Mark M' such that, OM' = 1 unit.
    ♦ Drop the perpendicular M'N' onto the y-axis.
• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM'}{OM}~=~\frac{ON'}{ON}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the y-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the y-coordinate of M' is $\small{\sin \left(\theta + \phi \right)}$ (Details here)
• So we get:
$\small{ON~=~A \left(\frac{\sin\left(\theta + \phi \right)}{1} \right)~=~A\,\sin \left(\theta + \phi \right)}$

• Therefore in this case, the vertical displacement of the sphere from O, at time 't' is given by:
$\small{x(t)~=~A \,\sin\left(\theta + \phi \right)}$

3. Let us draw the graph. It is shown in fig.14.14 below. For easy comparison, the previous graph in red color, is also shown as such. The new graph is shown in green color. For the new graph:
   ♦ A is the same 3 units
   ♦ T is the same 10 s
   ♦ $\small{\phi}$ is assumed to be $\small{\frac{\pi}{3}}$

Fig.14.14

• We get:
$\small{x(t)~=~A\,\sin\left(\frac{2 \pi t}{T}+\phi \right)~=~(3)\,\sin\left(\frac{2 \pi t}{10}+\frac{\pi}{3} \right)~=~(3)\,\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$

• Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~3~=~(3)\,\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$
$\small{\Rightarrow 1~=~\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 0.83, 10.83, 20.83, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 3

(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-3~=~(3)\,\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$
$\small{\Rightarrow -1~=~\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 5.83, 15.83, 25.83, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −3

(iii) To find the instants at which the red block reaches the equilibrium point, we need to solve the equation:
$\small{x(t)~=~0~=~(3)\,\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$
$\small{\Rightarrow 0~=~\sin\left(\frac{\pi t}{5}+\frac{\pi}{3} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 3.33, 8.33, 13.33, 18.33, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.

(iv) The green curve is an exact replica of the red curve. But the green is shifted by a small amount  towards the left.
• That means:
The green reaches the extreme points and zero points earlier than the red.
• For example:
   ♦ Red reaches a maximum at t = 2.5 s. Green reaches a maximum at t = 0.83 s
   ♦ Red reaches a minimum at t = 7.5 s. Green reaches a minimum at t = 5.83 s
   ♦ Red reaches a zero at t = 5 s. Green reaches a maximum at t = 3.33 s
• We say that:
The two oscillations are out of phase by an angle of $\small{\frac{\pi}{3}}$ radians
• The angle $\small{\phi~=~\frac{\pi}{3}}$
• $\small{\phi}$ is called the phase constant.

---

In the next section, we will see the oscillation of the simple pendulum.

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14.1 - Displacement of An Oscillating particle

In the previous section, we saw period and frequency. In this section, we will see displacement.

Let us first see the details of an oscillation. It can be written in 6 steps:
1. In fig.14.3 (a) below, a red block of mass m, is attached to a spring. The other end of the spring is fixed to a rigid wall. The block is resting on a friction-less surface.

Fig.14.9

2. In fig.14.3 (b), the block is pulled towards the right by a distance A. From this position, the block is released from rest. It will then travel towards the left. Even after traveling a distance 'A', it will continue to travel towards the left. That is., even after passing the point x= 0, it will continue to travel towards the left. This is shown in fig.c.

3. But once the point x = 0 is passed, the block will begin to experience a resistive force. The block is able to overcome the resistive force, and reach up to x = −A. Once it reaches x=−A, it stops. That is., it's velocity becomes zero. This is shown in fig.d.

4. Then it starts the reverse journey. In the reverse journey, even after passing the point x=0, it will continue to travel towards the right. This is shown in fig.e.

5. But once the point x = 0 is passed, the block will begin to experience a resistive force. The block is able to overcome the resistive force, and reach up to x = A. Once it reaches x=A, it stops. That is., it's velocity becomes zero. This is shown in fig.f.

6. At this point, one cycle is complete. Then it again starts the journey towards left. This process continues giving rise to continuous oscillation.

---

Displacement

The above 6 steps give us a basic understanding about oscillation. Now we will derive an expression for displacement. It can be done in 9 steps:
1. We saw that, the red block is oscillating between x=A and x=−A. Let us mark those two points as P and Q respectively, on the x-axis. This is shown in fig.14.4 below:

Fig.14.4

• The coordinates of P and Q are (A,0) and (−A,0) respectively. So O is the equilibrium position of the block.

2. Draw the red circle with center at O and radius equal to A.
• The magenta sphere is performing uniform circular motion along the red circle.
• The angular velocity of the magenta sphere is $\small{\omega}$.
• That means, the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.

3. Consider the instant when the magenta sphere is at M. At that instant, the line OM makes an angle $\small{\theta}$ with the x-axis.
• Drop the perpendicular MN from M, onto the x-axis.
• From the right triangle OMN, we get:
$\small{ON~=~ OM\,\cos \theta~=~A\,\cos\theta}$.

4. Now, ON is the horizontal displacement of the magenta sphere from the equilibrium position O. So we get a method to write the horizontal displacement of the magenta sphere from O.

◼ Let us check for other points. For that, we will try to write a general method which is applicable to all points.

(i) In fig.14.5(a) below, M is in the I quadrant.
    ♦ On OM, Mark M' such that, OM' = 1 unit.
    ♦ Drop the perpendicular M'N' onto the x-axis.

Fig.14.5

• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM}{OM'}~=~\frac{ON}{ON'}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$   
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the x-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the x-coordinate of M' is $\small{\cos \theta}$ (Details here)
• So we get:
$\small{ON~=~A \left(\frac{\cos \theta}{1} \right)~=~A\,\cos \theta}$

(ii) In fig.14.5(b) above, M is in the II quadrant.
    ♦ On OM, Mark M' such that, OM' = 1 unit.
    ♦ Drop the perpendicular M'N' onto the x-axis.
• OMN and OM'N' are similar triangles. So we can write:
$\small{\frac{OM}{OM'}~=~\frac{ON}{ON'}}$
$\small{\Rightarrow~ON~=~OM\left(\frac{ON'}{OM'} \right)}$   
$\small{\Rightarrow~ON~=~A \left(\frac{ON'}{OM'} \right)}$
• Length of ON' is same as the x-coordinate of M'.
• Since OM' = 1 unit, the point M' is on the unit circle, and so, the x-coordinate of M' is $\small{\cos \theta}$
• So we get:
$\small{ON~=~A \left(\frac{\cos \theta}{1} \right)~=~A\,\cos \theta}$
• Since M is in the II quadrant, $\small{\cos \theta}$ will be −ve. Indeed, N will have a −ve x-coordinate because it is on the −ve side of the x-axis.

(iii) In fig.14.5(c) above, M is in the III quadrant.
We can write similar steps and obtain the same result.

(iv) In fig.14.5(d) above, M is in the IV quadrant.
We can write similar steps and obtain the same result.

• So whichever be the quadrant, the horizontal displacement of the magenta sphere will be $\small{A\,\cos \theta}$ 

5. We see that, this method is very effective to write the horizontal displacement of the magenta sphere. But we want the horizontal displacement of the red block.
• So we assume that:
   ♦ At the instant when the block is released from P, the sphere starts the revolution from P
   ♦ At the instant when the block reaches O, the sphere reaches R
   ♦ At the instant when the block reaches Q, the sphere also reaches Q
   ♦ At the instant when the block returns through O, the sphere reaches S
   ♦ At the instant when the block returns back at P, the sphere also reaches back at P

6. That means, the time period (T) for one oscillation of the block is same as the time for one revolution of the sphere. Using this information, we can write an expression for $\small{\theta}$
• Time for one revolution of the sphere = T seconds
⇒ Time for $\small{2 \pi}$ radians = T seconds
⇒ Time for 1 radian = $\small{\frac{T}{2 \pi}}$ seconds
⇒ Angular distance covered in 1 second  = $\small{\frac{2 \pi}{T}}$ radian
• The stop-watch is turned on at the instant when the block is released from P. At that same instant, the sphere starts the revolution from the same point P.
• Consider the instant at which the reading in the stop-watch is t. Let at that instant, the sphere be at M.
• So when the reading is t, the angular distance covered is $\small{\theta}$
• Angular distance covered in 1 second  = $\small{\frac{2 \pi}{T}}$ radian
⇒ Angular distance covered in t seconds  = $\small{\frac{2 \pi t}{T}}$ radian
⇒ $\small{\theta}$  = $\small{\frac{2 \pi t}{T}}$ radian
• So we can write:
At any time t,
The horizontal displacement of the sphere from O
= The horizontal displacement of the block from O
= $\small{A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$
• That means:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$

7. We derived the equation: $\small{x(t)~=~A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)}$.
• If we want, we can use $\small{\omega}$ instead of $\small{T}$.
• We have:
the angle subtended by the sphere, at O, in each second, is $\small{\omega}$.
• So in $\small{t}$ seconds, the sphere will subtend $\small{\omega\,t}$ radians at O
• But the angle subtended by the sphere, at O, in $\small{t}$ seconds, is $\small{\theta}$.
• That means: $\small{\theta~=~\omega\,t}$
• Thus we get:
$\small{x(t)~=~A\,\cos\theta~=~A\,\cos\left(\frac{2 \pi t}{T} \right)~=~A\,\cos\left(\omega\,t \right)}$.

8. Here, the only variable on the R.H.S is $\small{t}$.
   ♦ $\small{t}$ is the independent variable
   ♦ $\small{x}$ is the dependent variable
• So while drawing the graph, we must plot $\small{t}$ along the x-axis and $\small{x}$ along the y-axis.

9. One such graph is shown in fig.14.6 below.
   ♦ A is assumed to be 2.5 units
   ♦ T is assumed to be 6 s

Fig.14.6

• We get:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T} \right)~=~(2.5)\,\cos\left(\frac{2 \pi t}{6} \right)~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$

• Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow 1~=~\cos\left(\frac{\pi t}{3} \right)}$

• This is a trigonometrical equation. We have seen the method to solve such equations, in our math classes (Details here).
• We get: t = 0, 6, 12, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 2.5

(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow -1~=~\cos\left(\frac{\pi t}{3} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 3, 9, 6, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −2.5

(iii) To find the instants at which the red block reaches the equilibrium position, we need to solve the equation:
$\small{x(t)~=~0~=~(2.5)\,\cos\left(\frac{\pi t}{3} \right)}$
$\small{\Rightarrow 0~=~\cos\left(\frac{\pi t}{3} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 1.5, 4.5, 7.5, 10.5, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.

(iv) We can obtain time period T from the graph:
• (6,2.5) and (12,2.5) are two consecutive +ve extreme points.
• The difference between their x-coordinates is (12 − 6) = 6 s
• This 6s is the time period T that we assumed to draw the graph.
• The red block can be assumed to start from the +ve extreme at t = 6s. It then travels to the −ve extreme and returns to the +ve extreme at t = 12 s. That means, the time for one complete cycle is 6 s 

---

Phase constant

This can be explained in 3 steps:
1. Consider fig.14.4 again. We turned the stop-watch on, when the red sphere was at P. Then we noted the time 't' at which the red sphere is at the arbitrary point M. The angle MOP at time 't' is denoted as '$\small{\theta}$'

2. The same fig.14.4 is modified and shown again in fig.14.7 below:

Fig.14.7

• Here, the stop-watch is turned on, when the sphere is at B. At B, the line OB already makes an angle $\small{\phi}$ with the x-axis.
• So at time 't', the sphere is at the arbitrary point M and the line OM makes an angle of $\small{\theta + \phi}$ with the x-axis
• Therefore in this case, the horizontal displacement of the sphere from O, at time 't' is given by:
$\small{x(t)~=~A \,\cos\left(\theta + \phi \right)}$

3. Let us draw the graph. It is shown in fig.14.8 below. For easy comparison, the previous graph in red color, is also shown as such. The new graph is shown in green color. For the new graph:
   ♦ A is the same 2.5 units
   ♦ T is the same 6 s
   ♦ $\small{\phi}$ is assumed to be $\small{\frac{\pi}{6}}$

Fig.14.8

• We get:
$\small{x(t)~=~A\,\cos\left(\frac{2 \pi t}{T}+\phi \right)~=~(2.5)\,\cos\left(\frac{2 \pi t}{6}+\frac{\pi}{6} \right)~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$

• Let us write some of the information that can be obtained from the graph:
(i) To find the instants at which the red block reaches the positive extreme, we need to solve the equation:
$\small{x(t)~=~2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow 1~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 5.5, 11.5, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all 2.5

(ii) To find the instants at which the red block reaches the negative extreme, we need to solve the equation:
$\small{x(t)~=~-2.5~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow -1~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 2.5, 8.5, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all −2.5

(iii) To find the instants at which the red block reaches the equilibrium point, we need to solve the equation:
$\small{x(t)~=~0~=~(2.5)\,\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$
$\small{\Rightarrow 0~=~\cos\left(\frac{\pi t}{3}+\frac{\pi}{6} \right)}$

• This is a trigonometrical equation. Solving it, we get:
t = 1, 4, 7, 10, . . .
• The points are marked on the graph. Indeed we see that, the y-coordinates at these points are all zero.

(iv) The green curve is an exact replica of the red curve. But the green is shifted by a small amount  towards the left.
• That means:
The green reaches the extreme points and zero points earlier than the red.
• For example:
   ♦ Red reaches a maximum at t = 6 s. Green reaches a maximum at t = 5.5 s
   ♦ Red reaches a minimum at t = 3 s. Green reaches a minimum at t = 2.5 s
   ♦ Red reaches a zero at t = 7.5 s. Green reaches a maximum at t = 7 s
• We say that:
The two oscillations are out of phase by an angle of $\small{\frac{\pi}{6}}$ radians
• The angle $\small{\phi~=~\frac{\pi}{6}}$
• $\small{\phi}$ is called the phase constant.

---

In the next section, we will see the oscillation of the same spring in the vertical direction.

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