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.

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

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Chapter 6.3 - Work done by a Force

In the previous section, we saw different properties of scalar products. In this section, we will see work done by a force and also some basics about kinetic energy.

We have seen some basics about kinetic energy in our high school classes. The following two links will give those details

High school physics - Kinetic energy lesson 1

High school physics - Kinetic energy lesson 2

It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion

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We will write our present discussion on kinetic energy in steps:
1. In chapter 3, we saw the third equation of motion:
$\mathbf\small{v^2-v_0^2=2ax}$
• v₀ is the initial velocity, v is the final velocity, a is the acceleration and x is the distance traveled
2. That means, an object was moving with an initial velocity of v₀ 
• It was given an acceleration of a
    ♦ By Newtons's second law, acceleration is achieved by the application of a force F
• As a result, it's velocity changed to v
• During the 'time interval in which this change in velocity occurred', the object traveled a distance x  
• This is shown in fig.6.6 below:

Fig.6.6

3. Let the mass of the object be m
• Multiplying both sides by ^m⁄₂, we get:
$\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=max}$
4. From Newton's second law, (m×a) = F
So we get: $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=Fx}$
5. We know that, to cause an acceleration, a force is required.
■ So it is obvious that, the 'F' in step (4) is the force which caused the acceleration 'a' which in turn changed the velocity from v₀ to v 
6. Let us analyze the equation in (4) in some detail:
• On the right side we have 'Fx'. It is Force × displacement
• But 'Force × displacement' is work done by the force
• So the right side of the equation in (4) gives the 'amount of work done by F'
7. Then the left side also must be 'work done'
■ How is that possible?
• It is simple actually. The work done by F is stored in the object as 'energy'
• The second term on the left side in (4) is $\mathbf\small{\frac{1}{2}mv_0^2}$ 
    ♦ This is the energy which the object posessed when it was moving with velocity v₀.
• The first term on the left side is $\mathbf\small{\frac{1}{2}mv^2}$     
    ♦ This is the energy which the object is possessing when it is moving moving with velocity v.
8. The difference $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2}$ is the 'change in energy' acquired by the object
• The object was able to acquire this 'change in energy' because, work was done on it by F
• The 'change in energy' acquired is exactly equal to the 'work done by F'. Hence we have the '=' sign between the LHS and RHS in (4)
9. $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial energy 
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final energy
■ The energy posessed by an object due to it's motion is called kinetic energy
So we can write:
• $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial kinetic energy
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final kinetic energy
■ If an object of mass 'm' is moving with a velocity 'v', then the kinetic energy pocessed by that object is $\mathbf\small{\frac{1}{2}mv^2}$  
10. We see many examples of kinetic energy in our day to day life
• A 'pebble A' moving with a certain velocity possesses kinetic energy. It can hit another 'pebble B' and displace it.
• If 'pebble A' is stationary, it does not possess kinetic energy. It will not be able to displace 'pebble B'

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The work-energy (WE) theorem states that: The change in kinetic energy of a particle is equal to the work done on it by the net force

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Solved example 6.4

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height of 1 km. It hits the ground with a speed of 50.0 m s^-1. (a) What is the work done by the gravitational force? (b) What is the work done by the unknown resistive force?
Solution:
1. Rain drop is formed in the upper atmosphere when the surrounding temperature falls. The water vapor then condenses to form liquid water droplets. These water droplets combine together to form a larger water drop. Once a drop is formed, gravity will pull it towards the earth. 
• So we see that the drop is in free fall. And it's initial velocity v₀ is zero
2. The drop in our problem falls from a height of 1 km (1000 m)
• We know that velocity attained during free fall do not depend on mass. It depends only on the height of fall
• So how much velocity will be attained by the drop when it reaches the ground?
• We can find this in two steps:
(i) We have x = v₀t + ¹⁄₂at²
• Entering known values, we get: 1000 = 0 + ¹⁄₂ × 10 × t²
• So we get t = √200 = 10√2 seconds
(ii) We have v = v₀ + at
• Entering known values, we get: v = 0 + 10 × 10√2 = 100√2 ms^-1 
3. But it is given that, the drop hits the ground with a velocity of 50 ms^-1.
• That means the drop was not able to attain 100√2 ms^-1.
• This is because the air resistance acted in the upward direction thus opposing the motion of the drop
• So two forces acted on the drop
(i) The downward gravitational force
(ii) The upward resistive force
4. When a force act on an object and moves it through a certain distance, work is being done by the force on that object
• This work will be stored in the object as kinetic energy
• So the gravitational force will give the drop a kinetic energy equal to ¹⁄₂mv² 
= ¹⁄₂ × 0.001 × (100√2)² = 10 joules
5. But this much kinetic energy was not attained. The velocity attained was only 50 ms^-1.
• So actual kinetic energy attained = ¹⁄₂ × 0.001 × (50)² = 1.25 joules      
6. The difference occurred because, the air resistance did some work in the opposite direction
• We can write:
Work done by gravity - Work done by air resistance = Actual kinetic energy attained
• Entering known values, we get: 10 - Work done by air resistance = 1.25
• So 'Work done by air resistance' = (10-1.25) = 8.75 joules
• Note: After discussing about 'potential energy', we will do this problem again using those concepts also. 

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• In the discussions so far in this section, the motions were all rectilinear (that is., one dimensional)

• In one dimensional motion, we do not need to use vector notations
• Next we will consider two dimensional motion
1. Consider fig.6.7 below:

Fig.6.7

• A force $\mathbf\small{\vec{F}}$ is acting at an angle θ to the horizontal 

2. Because of this angle, the block will tend to move in two directions:
(i) horizontally to the right  
(ii) vertically towards bottom
• In this case, there will not be a motion downwards. This is due to the presence of the floor.
• Even then, because of the slope of the force, it is a case of two dimensional motion and we have to use vector notations
3. Because of the slope of the force, there will be two force components
• One vertical and the other horizontal
• The horizontal component causes the block to move horizontally
• The vertical component is unable to cause any movement in the vertical direction because of the floor
4. So what is the work done?
• We see that the displacement is in the horizontal direction
• Word done is (Force × displacement)
• But to use this equation, the force must be in the same direction as the displacement
5. So we must take the horizontal component of the force
• Magnitude of the horizontal component is $\mathbf\small{|\vec{F}|\cos \theta}$
6. If the displacement vector is $\mathbf\small{\vec{d_x}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_x}|}$
7. So work done = $\mathbf\small{|\vec{F}|\cos \theta \times |\vec{d_x}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_x}|\cos \theta }$
• Note that, θ is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$    
■ Thus we can write: 
Work done = $\mathbf\small{\vec{F}.\vec{d_x}}$
That is., work done is the dot product of force vector and displacement vector.
8. In our present case, since there is no displacement in the vertical direction, there is no work done by the vertical component of the force
• But if the floor was absent, there would be work done in vertical direction also
• Can we use dot product to find that work? 
9. Consider fig.6.8 below:

Fig.6.8

• Since the floor is absent, the object moves both in horizontally and vertically
• We want the work done by the vertical component of $\mathbf\small{\vec{F}}$
• $\mathbf\small{\vec{F}}$ makes an angle θ₁ with the vertical
• So magnitude of the vertical component of $\mathbf\small{\vec{F}}$ = $\mathbf\small{|\vec{F}|\cos \theta_1}$
10. If the displacement vector in the vertical direction is $\mathbf\small{\vec{d_y}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_y}|}$
11. So work done = $\mathbf\small{|\vec{F}|\cos \theta_1 \times |\vec{d_y}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_y}|\cos \theta_1 }$
• Note that, θ₁ is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ Thus we can write:
In the vertical direction also, work done is the dot product
12. When ever we take the product of $\mathbf\small{|\vec{F}|}$ and cos θ, we are getting the component of that force along the direction of displacement
■ So we can write:
Work done by a force is the product of two items:
(i) Component of the force in the direction of displacement
(ii) Magnitude of the displacement
■ In equation form, we get:
Eq.6.15:. $\mathbf\small{W=\vec{F}.\vec{d}}$
From Eq.6.15, it is clear that, if displacement is zero, work done is zero

An example: 
• If a person push hard against a rigid wall, there is no displacement for the wall. 
    ♦ Then the work done by the person is zero
• The person will get tired after some time. This is due to the alternate contracting and relaxing of the muscles. And also due to the usage of internal energy.
• So the meaning of work in physics is different from it’s usage in everyday language.
Another example:
• A weight lifter holding a 150 kg mass steadily for 30 s does no work on that mass 
    ♦ This is because there is no displacement for that mass
Another example:
• A block may move large distance on a smooth floor. 
• This movement will not require any force since there is no friction to overcome. 
• In such a motion, displacement is large. 
• But force is zero. 
    ♦ Then work done is also zero
An interesting example:
• We have seen that gravitational force does work on a rain drop. 
• Now consider this: A block of mass m is moving horizontally on a smooth floor. 
• The gravitational force mg is acting continuously on the block. 
• So there is both force and displacement. 
• But the force is vertical and displacement is horizontal. 
    ♦ The angle between the force vector and displacement vector is 90^o 
• Then the dot product becomes zero because cos 90 = 0
• Thus in this case, we can say: the gravitational force does not do any work on the block
Another example:
• Assume that the moon’s orbit around the earth is a perfect circle
• Then the gravitational force between the earth and the moon will be in the radial direction
• The motion of the moon at any instant is tangential to the orbit
• We know that in any circle, the radius is perpendicular to tangent
• This is shown in fig.6.9(a) below:

Fig.6.9

• So the direction of gravitational force is perpendicular to the direction of displacement of the moon
• Thus we can say: the work done by gravitational force on the moon is zero

Negative work

1. Consider the block in fig.6.9(b) above. The displacement of the block is horizontal and towards the right. 
• But the force is acting in a sloping direction towards the left
2. In this case also, to find the work done, we take the same dot product: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$  
• But here θ is greater than 90^o 
• From trigonometry classes we know that when θ is between 90^o and 180^o, cos θ becomes negative.
    ♦ For example, cos 150 = -0.8660
• So the final dot product becomes negative
■ That means the work done is negative
• This is understandable because force is acting in a direction opposite to that of the displacement.

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Fig.6.9(b) above can be used to discuss a few more details about the 'influence of θ on work done'
We will write the steps:
1. In the fig.6.9(b), when θ = 0, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the right
• When θ = 0, cos θ = 1 
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the maximum possible value
■ That means  when θ = 0, the 'work done' will be the maximum
■ In general, when θ = 0, the dot product of two vectors will be having the maximum possible value
2. In the fig.6.9(b), when θ is greater than zero but less than 90, $\mathbf\small{\vec{F}}$ will be sloping towards the right
• When θ is greater than zero but less than 90, cos θ will be a 'positive fraction'
    ♦ For example, cos 40 = 0.7660
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'positive fraction'  
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be positive but less than maximum
3. In the fig.6.9(b), when θ = 180, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the left
• When θ = 180, cos θ = -1 
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '-1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the 'maximum possible magnitude', but in the opposite direction
■ The work done is opposite to the direction of motion. Such work is considered as 'negative work'
• The work done by friction is an example of this case. On a horizontal surface, the angle between the displacement vector and the 'frictional force vector' will be 180^o.
■ In general, when θ = 180, the dot product of two vectors will be having the least possible value because of the 'negative sign'
4. In the fig.6.9(b), when θ is greater than 90 but less than 180, $\mathbf\small{\vec{F}}$ will be sloping towards the left
• When θ is greater than 90 but less than 180, cos θ will be a 'negative fraction'
    ♦ For example, cos 135 = -0.7071
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'negative fraction'  
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be negative but greater than the least work in (3)

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Dimensions of work

1. We have: work = Force × displacement
2. Force = Mass × acceleration
3. Acceleration = change in velocity per second
4. velocity = distance/time
• So working on the above steps from velocity upwards, we get:
1. Velocity has dimensions:  [LT^-1]
2. Acceleration has dimensions:  [LT^-2]
3. Force has dimensions:  [MLT^-2]
4. Work has dimensions: [ML²T^-2]

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Solved example 6.5

A cyclist comes to a skidding stop in 10 m. During this process the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Solution:
• The cyclist stops pedaling and applies the brakes
• If he was pedaling at a normal speed, he could bring the cycle to a stop without skidding  
• If the speed was large, and brakes are applied suddenly, wheels would stop spinning instantly and the cycle would skid. The cyclist in our problem encountered this situation
• Now we can write the steps:
Part (a): 
1. When skidding occurs, road applies frictional force on the cycle
• It is given that the magnitude of this force $\mathbf\small{\vec{F}}$  is 200 N
2. Also it is given that the force acted for a distance of 10 m
• So magnitude of the displacement vector $\mathbf\small{\vec{d}}$ is 10 m
3. If we assume that the displacement vector to be from left to right, then force vector will be from right to left
• This is because, the frictional force is always opposite to the direction of motion
• Thus the angle θ between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is 180^o   
4. So the magnitude of the work done = $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$
= 200 × 10 × cos 180 
= 200 × 10 × -1 
= -2000 J
5. It is this negative energy that brings the cycle to a stop
• Let us apply the work-energy theorem to this situation:

■ The work-energy (WE) theorem states that: The change in kinetic energy of a particle is equal to the work done on it by the net force
• The change in kinetic energy = k₂ -k₁
• The final kinetic energy k₂ is zero
• So the change in kinetic energy is negative
• That means the work done by the net force (frictional force) is negative
Part (b):
1. The road applies a force of 200 N on the cycle
2. By Newton's third law, the cycle will apply a force of 200 N on the road
3. But the road does not undergo any displacement
• So work done by the cycle on the road is zero
4. We can write:
• According to Newton's third law, the following two forces will be always equal and opposite:
(i) Force applied by a body A on another body B
(ii) The reaction force applied by B on A 
• But the following works need not be equal and opposite:
(i) Work done by a body A on another body B 
(ii) Work done by the 'reaction force of B' on A

Solved examples 6.6
A force $\mathbf\small{\vec{F}=2\hat{i}+3\hat{j}+4\hat{k}}$ N causes displacement of an object. The position vector of the initial position of the object is $\mathbf\small{\vec{r_1}=2\hat{i}+3\hat{j}+1\hat{k}}$ m. The position vector of the final position is $\mathbf\small{\vec{r_2}=\hat{i}+\hat{j}+\hat{k}}$ m. What is the work done by $\mathbf\small{\vec{F}}$?
Solution:
1. Displacement vector $\mathbf\small{\vec{d}=\vec{r_2}-\vec{r_1}}$ (Details here)    
• Thus we get: $\mathbf\small{\vec{d}=(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})=(-\hat{i}-2\hat{j})}$ m
2. Work done = $\mathbf\small{\vec{F}.\vec{d}=F_xd_x+F_yd_y+F_zd_z}$ (Using Eq.6.8)
= (2 × -1 + 3 × -2) = (-2-6) = -8 J

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In the next section, we will see more details about kinetic energy

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