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 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 

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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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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.

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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 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 

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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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Chapter 14 - Oscillations

In the previous section, we saw specific heats of gases. In this chapter, we will see oscillations.

• In some earlier chapters, we have seen rectilinear motion, projectile motion and uniform circular motion.
• In uniform circular motion, a particle takes the same time to complete each revolution. So it is a periodic motion.
• In this chapter, we will see another type of periodic motion, called oscillation.
• In oscillation, a particle moves to and fro about a mean position.
   ♦ Motion of the pendulum is an example
   ♦ Motion of the piston inside the cylinder of an engine is another example.   
• To learn about periodic motion and oscillatory motion, we must first see some fundamental concepts like period, frequency, displacement etc., They are explained below:

Periodic and Oscillatory motions

Some basics can be written in 5 steps:
1. Consider an insect climbing up a wall. It starts climbing from the ground level. It travels with a uniform speed u₁. In a time duration of t₁, it reaches a height h. This is shown by the red line in fig.14.1(a) below:

Fig.14.1

• After t₁, the insect is not able to climb upwards. Neither is it able to hold on. So it climbs down to the ground with a uniform speed of u₂.  The time taken for reaching the ground is t₂. This is shown by the green line in fig.14.1(a) above.
• Note that, t₁ is larger than t₂. This is because, time taken to reach the ground is smaller.
• The sum (t₁ + t₂) is T. This T is the time required to start from  the ground and reach back to the ground. If the insect keep repeating the "climbing and falling" process, we will get several repetitions in the graph. The ground is then considered as the mean position.
• The red line is familiar to us. We have plotted many distance-time (s-t) graphs in rectilinear motion. If the motion is uniform, we get the same red line. It is the plot of the equation s = ut. In our present case, u₁ is the slope of the red line. u₁ is constant.
• The green line can also be given a similar explanation. It has a negative slope because, velocity is negative for downward motion. So slope of the green line is −u₂.
• In fig.14.1, instead of s, we plot x(t) along the y-axis. This is because, in periodic motion, the displacement from the mean position is denoted by x. It is a function of time. So we write x(t).

2. Consider the game of bouncing the ball between palm and ground.
• The game starts when the ball is thrown to the ground from a height h. It is thrown with an initial speed of u₁. Since the ball is under the influence of gravity, speed is not uniform.
• The displacement at any instant 't' is given by:
$\small{x(t)\,=\,u_1 t + \frac{1}{2} a t^2}$. This is the equation of a parabola.
• In a time duration of t₁, it reaches the ground. This is shown by the red parabola in fig.14.1(b) above.
• After t₁, the ball bounces upwards with a speed of u₂. Since the ball is under the influence of gravity, speed is not uniform. The displacement at any instant 't' is given by:
$\small{x(t)\,=\,u_2 t - \frac{1}{2} a t^2}$. This is the equation of a parabola.
• In a time duration of t₂, it reaches the palm. This is shown by the  green parabola in fig.14.1(b) above.
• The sum (t₁ + t₂) is T. This T is the time required to start from  the palm and reach back to the palm. As the game continues, we will get several repetitions in the graph. The ground is then considered as the mean position.
• Note that, the red and green parabolas are graphs. They are not the paths followed by the ball. The path of the ball is always vertical. While playing this game, the ball cannot be thrown in a parabolic path. If the ball is thrown in a parabolic path, it will not bounce back to the palm. It will bounce away from the player.

3. A motion that repeats itself at regular intervals of time is called a periodic motion.

4. Consider the red sphere in fig.14.2 below. It is at the bottom of the yellow bowl.

Fig.14.2

• In this position, the sphere is at equilibrium. If the sphere is left alone, it will remain there forever.
• But if the sphere is given a displacement, a force will come into play, which tries to bring the sphere back to the equilibrium position. As a result, the sphere will perform oscillation or vibration in the bowl.
• Oscillating bodies eventually come to rest. In our present case, the sphere will come to rest at the bottom of the bowl.
• This is due to the friction between the sphere and the bowl. The surrounding air also causes the sphere to come to rest. This type of oscillation is called damped oscillation.
• However, an oscillation can be forced to remain oscillating, by providing an external agency. This type of oscillation is called forced oscillation. We will see the details of both types in later sections of this chapter.

5. Every oscillatory motion is a periodic motion.
• But every periodic motion need not be an oscillatory motion.
   ♦ Circular motion is periodic but not oscillatory.

Period and frequency

This can be explained in 7 steps:
1. Consider the earlier fig.14.1(a). We turn on the stop-watch at the instant when the insect begins to climb up. When the reading in the stop-watch is T, the insect is back on the ground, and the next cycle begins.

2. In fig.14.1(b), we turn on the stop-watch at the instant when the ball is thrown. When the reading in the stop-watch is T, the ball is back at the palm, and the next cycle begins.

3. So we can write:
The smallest duration of time after which the next cycle begins, is called the period of the periodic motion. It is denoted by the symbol T.
• The S.I unit of period is second.

4. For periodic motions which are too fast, period will be very small. In such cases, instead of s, we can use other units.
• For example, the period of vibration of a quartz crystal is expressed in microseconds.
   ♦ One microsecond is $\small{10^{-6}}$ seconds.
   ♦ Microsecond is abbreviated as: $\small{{\rm{\mu s}}}$

5. For periodic motions which are too slow, period will be very large. In such cases also, instead of s, we can use other units.
• For example, the orbital period of some  planets are expressed in earth days.
   ♦ For mercury, it is 88 earth days.
   ♦ One earth day is 24 hours.

6. Suppose that, period of a periodic motion is 30 s. Then in one minute, two cycles will occur.
   ♦ That is., in 60 s, two cycles will occur
   ♦ That is., in 1 s, 2/60 cycles will occur.
   ♦ That is., in 1 s, 1/30 cycles will occur.
   ♦ That is., in 1 s, 1/30 of one cycle will occur.
• The number of cycles in one second is called frequency. It is represented by the symbol $\small{\nu}$. In our present case, $\small{\nu}$ = 1/30.
• Note that, 1/30 is the reciprocal of 30. And 30 is the period in seconds.
• So we can write:
Frequency is the reciprocal of period. The period must be in seconds.
   ♦ That is., $\small{\nu~=~\frac{1}{T}}$
   ♦ where T is n seconds.
• Since we are taking the reciprocal of "period in seconds", the unit of frequency is $\small{{\rm{s^{-1}}}}$.
This $\small{{\rm{s^{-1}}}}$ is given a special name. It is called hertz (abbreviated as Hz), in honor of the scientist Heinrich Rudolph Hertz, who discovered radio waves.
• We can write:
1 hertz = 1 Hz = 1 cycle per second = 1 $\small{{\rm{s^{-1}}}}$.

7. Note that, $\small{\nu}$ need not be an integer. It can be a decimal or a fraction.

Let us see a solved example:

Solved example 14.1
On an average, a human heart is found to beat 75 times in a minute. Calculate it's frequency and period.
Solution:
1. First, let us calculate the frequency:
Number of occurrences in one minute = 75
⇒ Number of occurrences in 60 s = 75
⇒ Number of occurrences in 1 s = 75/60
⇒ Frequency $\small{\nu~=~\frac{75}{60}~=~\frac{5}{4}~=~1.25~{\rm{s^{-1}}}}$
2. Now we can calculate the period:
We know that, frequency is the reciprocal of period. It follows that, period is the reciprocal of frequency. So we can write:
Period $\small{T~=~\frac{1}{\nu}~=~\frac{4}{5}~=~0.8 {\rm{s}}}$

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In the next section, we will see displacement.

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Chapter 8.18 - Geostationary Satellite

In the previous section, we completed a discussion on energy of orbiting earth satellites
• In this section we will see Geostationary satellites

1. In fig.8.51 below, the earth is shown as a blue sphere

Fig.8.51

• The axis of rotation of the earth is shown in yellow color
    ♦ It passes through the N and S poles
• The direction of rotation is indicated by the orange curved arrow
2. A satellite shown in red color orbits around the earth
• The orbit of the satellite is circular in shape. It is shown in green color
3. In fig.8.52 below, the earth is shown to be transparent
• So center of the earth is visible. The center is indicated by a small white sphere

Fig.8.52

4. A magenta line is drawn from the center of earth to center of the satellite
• This magenta line intersects the surface of the earth at P
5. Suppose that, the 'time period T_E of the earth (which is 24 hours)' and 'time period T_S of the satellite' are the same
• That is., both earth and the satellite complete one rotation at the same time
• Then, both of them will be rotating as a single unit
• The point P will not change
• For a person standing at P, the satellite will appear to be stationary
6. Not only for a person at P. In fig.8.53 below, two more magenta lines are drawn in random directions

Fig.8.53

• Those new lines intersect the surface of the earth at P₁ and P₂
• For those standing at P₁ and P₂ also, the satellite will appear to be stationary
• In fact, since the earth and the satellite rotate as a single unit, the satellite will appear to be stationary for every person on earth
• Such a satellite is called a geostationary satellite
7. One important point has to be noted:
■ For a satellite to be geostationary, it's orbit must lie in the equatorial plane
• This can be explained in 9 steps:
(i) Let us assume that orbits of all the earth satellites are circles
(ii) The centers of all those circles must coincide with the center of the earth
• This is because, the gravitational force (which provides the centripetal force), acts towards the center of the earth
(iii) So we can have the green orbits shown in fig.8.54 below. Both of them have their centers at the center of the earth

Fig.8.54

• But the magenta orbit is not possible because, it's center is not at the center of the earth
(iv) The view in fig.8.55 below helps to understand the difference between the three orbits

Fig.8.55

We see that:
• The green orbits has centers at the earth's center
    ♦ So the green orbits are possible
• The center of the magenta orbit is away from the earth's center
    ♦ So the magenta orbit is not possible for satellites
(v) Out of the two green orbits, one is horizontal and the other is inclined
• Consider the horizontal green orbit
    ♦ We know that, center of all green orbits coincide with the center of the earth
    ♦ So the horizontal green orbit falls exactly on the equatorial plane
(We know that, the equator is a circle. The plane in which this circle lies, is the equatorial plane)
(vi) For a satellite to be geostationary, it's orbit should lie in the equatorial plane
• Let us see what happens if another green orbit is selected:
(vii) In the fig.8.56 below, the green orbit is inclined. It does not lie in the equatorial plane

Fig.8.56

• P is the familiar 'point of intersection of the magenta line' that we saw in the fig.8.52
(viii) A person standing initially at P will be moving along the horizontal red circle
• But the point P will be moving along the inclined yellow circle
• So the satellite will not appear to be stationary
(ix) Thus it is clear that, for a satellite to be geostationary, it's orbit must lie in the equatorial plane
• Let the time period T_S of the satellite in fig.8.56 be 24 hours
• Then satellite will reach the exact same spot after 24 hours. But it will not appear to be stationary for a viewer on earth
• Such satellites are called geosynchronous satellites
• All geostationary satellites are geosynchronous satellites. But all geosynchronous satellites are not geostationary satellites
8. Now we can write the definition for geostationary satellites:
Satellites in circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary satellites
9. Now we will see some mathematical calculations related to such satellites:
• We have Eq.8.26 that we derived in a previous section: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E+h)^{3}}{{G\,M_E}}}}$
$\mathbf\small{\Rightarrow (R_E+h)^{3}={\frac{T^2\,G\,M_E}{4\pi^2}}}$
$\mathbf\small{\Rightarrow (R_E+h)=\left({\frac{T^2\,G\,M_E}{4\pi^2}}\right)^{1/3}}$
10. In the above equation, only h and T are the variables
• If we put T = 24 hours (converted into seconds), we will get: h = 35800 km
11. This is a very large height. Powerful rockets will be required to take the satellites to such heights
But having satellites at such heights have many practical advantages. Let us see some of them. We will write them in steps:
(i) The ionosphere is a particular layer in the earths atmosphere
• It's level above the surface of the earth is indicated by the red line in fig.8.57 below:

Fig.8.57

(ii) Antenna A is transmitting radio signals. Those signals are indicated by the blue arrows
• The radio signals get reflected by the ionosphere. So they can be received on a larger area on the earth's surface
(iii) Antenna B is transmitting television signals. Those signals are indicated by the yellow arrow
• The television signals have a higher frequency than radio signals
• Such high frequency signals are not reflected by the ionosphere
• So such signals cannot be received over a large area
(iv) It is clear that, the curvature of the earth's surface is playing a major role here
• If the surface was flat, we would be able to transmit the TV signals to a wider area, even if there is no reflection from the ionosphere
• But as seen from the fig., the opposite side of the curvature does not receive any signals
(v) If we put a geostationary satellite up in the sky, the signals can be picked up and send to the opposite side of the curvature. It is shown in fig.8.58(a) below:

Fig.7.58

• Note that, the satellite must be a geostationary one. Otherwise, the antenna B will have to continuously change the direction of transmission

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• In the next section we will see Polar satellites

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Chapter 5.4 - Relation between Time and Force

In the previous section we obtained the relation between force and 'change in velocity (v₂-v₁)'. 
• We saw that force is directly proportional to '(v₂-v₁)'. 
• In this section, we will see the relation between time and velocity.

■ In the earlier section, we obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ In the previous section, we obtained F ∝ (v₂-v₁) by:
• Changing velocities
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and t by:
• Changing t
• Keeping m and v constant
We will do a series of experiments here also:

Experiment 13:
1. Consider a car on a level road. See fig.5.8(a) below:

Fig.5.8

• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F₁₃ be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The car will not move at the same instant when F₁₃ is applied. It will take some time to start moving
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F₁₃ is applied
(ii) The instant when the car attains a velocity of say 2 ms^-1 
(we can fix any convenient value for the velocity. 2 ms^-1 is only an example)
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the speedometer reading is 2 ms^-1
• Let t₂ = 12
• Then Δt = (t₂ - t₁) = (5-0) = 12 s
7. So we can write:
• A force F₁₃ is required to push a car from rest and to move it with a velocity of 2 ms^-1

• The time required for this velocity change (from zero to 2 ms^-1) is 12 s

Experiment 14:
1. Consider the same car on the same road. See fig.5.8(b)
• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The car will not move at the same instant when F is applied. It will take some time to start moving
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F is applied
(ii) The instant when the car attains the same velocity in experiment 13. That is 2 ms^-1 
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the speedometer reading is 2 ms^-1
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 13.
• It is the 'time' that we want to change   
• That is., we want to attain the velocity 2 ms^-1 'in a shorter time' than the time in experiment 13
Let us fix it as 8 s
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the car to rest and start pushing it again 
8. We do the trials until the following two conditions are satisfied
(i) The car attains a velocity of 2 ms^-1.
(ii) This velocity is attained in a time duration of 8 s
• The force F which satisfies the both two conditions can be noted down as F₁₄
■ We can write:
• A force F₁₄ is required to push a car from rest, to move it with a velocity of 2 ms^-1.
• The time required for this velocity change (from zero to 2 ms^-1) is 8 s

---

• The experiment 14 is over

Now we make a comparison between the results of the two experiments 

■ We will find that F₁₃ < F₁₄.
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 13
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously the force required will be more
■ We can write: When 'time' decreases, greater force is required. 
Let us do another set of two experiments to confirm this:

Experiment 15:
1. Consider a 'trolley carrying a mass' on a level road. See fig.5.9(a) below

Fig.5.9

• It is in a state of uniform motion towards the right.
• It's velocity is 2 ms^-1. 
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F₁₅ be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The trolley will not stop at the same instant when F₁₅ is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F₁₅ is applied
(ii) The instant when the trolley attains a velocity of 0 ms^-1 
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the trolley comes to rest
• Let t₂ = 5
• Then Δt = (t₂ - t₁) = (3-0) = 5 s
7. So we can write:
• A force F₁₅ is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms^-1 to zero) is 5 s

Experiment 16:
• We repeat the experiment with the same trolley used in the previous experiment 15. 
• The mass contained inside it should not change.
• The following points should be noted:
    ♦ The trolley must be moving with the same velocity (2 ms^-1) as in experiment 15
    ♦ We want to bring the trolley to rest 
    ♦ We want to achieve this 'velocity change' (from 2 to zero) within a shorter time than in experiment 15. Let us fix it at 3 s

• We will write the steps:
1. Consider the 'trolley with the same mass' on the same floor as in experiment 15
• It is in a state of uniform motion.
• It's velocity is 2 ms^-1. 
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way. 
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't₁' in the stop-watch will be 0 
4. The trolley will not stop at the same instant when F is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following two instances:
(i) The instant when force F is applied
(ii) The instant when the trolley attains a velocity of 0 ms^-1 
6. To measure Δt, we must carefully note down the stop-watch reading 't₂' at the instant when the trolley comes to rest
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 15. 
• Also, we want to attain the velocity 0 ms^-1 in a lesser interval '3 s'
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 2 ms^-1' and try to stop it again 
8. We do the trials until the following two conditions are satisfied
(i) The trolley attains a velocity of 0 ms^-1.
(ii) This velocity is attained in 3 s
• The force F which satisfies the both two conditions can be noted down as F₁₆
■ We can write:
• A force F₁₆ is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms^-1 to zero) is 3 s, which is less than the 5 s obtained in experiment 15

---

• The experiment 16 is over

Now we make a comparison between the results of the two experiments 

■ We will find that F₁₅ < F₁₆
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 15
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously, the force required will be more
■ We can write: When 'time' decreases, greater force is required. 
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:

Experiment 17:
1. Drop a small stone from the top of a building 
2. Let a person standing at the foot of the building catch it 
3. The following points should be noted while catching:
    ♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
    ♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F₁₇ 
Experiment 18:
1. Drop the same stone which was used in experiment 17
• Also, the drop must be made from the same building, and from the same height in experiment 17
2. Let the person standing at the foot of the building catch it 
3. The following points should be noted this time also:
    ♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
    ♦ The person must lower his hands gradually while making the catch
    ♦ This method of catching will increase the 'time of stopping'
4. Let the force experienced by the person be F₁₈.

---

• Let us make a comparison between the results of experiments 17 and 18

■ It will be found that F₁₈ < F₁₇
• The following points may be noted:
(i) The stone was dropped from the same height in both experiments
    ♦ So the final velocity is the same in both the cases
(ii) The same stone was dropped in both the experiments
    ♦ So the mass remains the same
(iii) The person lower his hands gradually in experiment 18
    ♦ So more time was allowed to stop the stone
■ We can say
• We wanted to obtain the same effect as in experiment 17
• Greater time was allowed to obtain the same effect
• So the force required will be less

---

■ So we completed three sets of experiments. We can now write with confidence:

• When 'time' decreases, greater force is required 
• When 'time' increases, lesser force is required
In other words:
■ Force is inversely proportional to 'time'
• symbolically we write this as: F∝ ¹⁄_t
• The symbol '∝' stands for 'is proportional to'

---

• Thus we found the relation between 'time' and force.

• In the next section we will combine all the information together

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