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

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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 the sum of two periodic functions.

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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 

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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.

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

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