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.
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x(t)} & {~=~} &{ A\cos(\omega t)}
\\ {~\color{magenta} 2 } &{} &{v(t)} & {~=~} &{-\omega A\sin(\omega t)}
\\ {~\color{magenta} 3 } &{} &{a(t)} & {~=~} &{-\omega^2 A\cos(\omega t)}
\\ \end{array}}$
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{a(1.5)} & {~=~} &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi (1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{139.58{\rm{(m/s^2)}}}
\\ \end{array}}$
In the next
section, we will see the force law for SHM.
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