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.

---

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

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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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