14.4 - Simple Harmonic Motion

In the previous section, we saw that displacement from the equilibrium position can be specified using sine or cosine functions. In this section, we will see simple harmonic motion.

1. Consider a particle subjected to oscillation along the x-axis. Suppose that, the oscillation satisfies the following conditions:
(i) -`A and A are the extreme points.
(ii) The origin O is the equilibrium point.
(iii) Displacement from the origin can be specified using a sinusoidal function.
(Sine and cosine functions are examples for sinusoidal functions)
2. We have seen that the particle attached to a horizontal spring can satisfy the above three conditions. (see section 14.1)
Let us observe this oscillation. Assume that, the stop-watch is turned on when the particle is at O
3. Then, in addition to those three conditions, the particle satisfies more conditions:
(iv) When $\small{t~=~0}$, the particle is at O, and the velocity is maximum.
(v) When $\small{t~=~\frac{T}{4}}$, the particle is at +A, and the velocity is zero.
(vi) When $\small{t~=~\frac{T}{2}}$, the particle is at O, and the velocity is maximum.
(vii) When $\small{t~=~\frac{3T}{4}}$, the particle is at -`A, and the velocity is zero.
(viii) When $\small{t~=~T}$, the particle is at O, and the velocity is maximum.
(ix) When $\small{t~=~\frac{5T}{4}}$, the particle is at +A, and the velocity is zero.
so on . . .
4. If the nine conditions are satisfied, then that oscillation is called a simple harmonic motion. It is abbreviated as SHM.
5. In the previous sections, we became familiar with the function:
$\small{x(t)~=~A \cos\left(\omega t \,+\,\phi \right)}$
    ♦ $\small{x(t)}$ is the displacement $\small{x}$ as a function of time $\small{t}$
    ♦ $\small{A}$ is the amplitude
    ♦ $\small{\omega}$ is the angular frequency
    ♦ $\small{\omega t \,+\,\phi}$ is the phase. It depends on time.
    ♦ $\small{\phi}$ is the phase constant
• This is a sinusoidal function which can represent a SHM.

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Let us see some examples:

Example 1:
In fig.14.21 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.21

From the graph, we get two information:
(i) Information about period T
• Both red and green reach maximums at the same instants.
• Both red and green reach minimums at the same instants
• Both red and green reach zero at the same instants
• So red and green have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$
Green: $\small{x(t)~=~3 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The instants corresponding to maximums of red, can be obtained by solving the equation:
$\small{4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$ 

The instants corresponding to minimums of red, can be obtained by solving the equation:
$\small{-4~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The instants corresponding to zeros of red, can be obtained by solving the equation:
$\small{0~=~4 \cos\left(\frac{\pi t}{6} \,+\,\frac{\pi}{3} \right)}$

The same procedure can be used for green.

Always remember the relation between T and $\small{\omega}$, which is: $\small{\omega~=~\frac{2 \pi}{T}}$

So for the red, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\omega t}    & {~=~}    &{\frac{\pi t}{6}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\omega}    & {~=~}    &{\frac{\pi}{6}~=~\frac{2 \pi}{T}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{T}    & {~=~}    &{12~\text{seconds}}    \\
\end{array}}$

In the same way, we can find the period of green also.

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of red is 4 units.  
• The two magenta horizontal dashed lines indicate that, the amplitude of green is 3 units.

◼ Based on this example, we can write:
Two simple harmonic motions may have the same phase $\small{\left(\omega t \,+\,\phi \right)}$. But they can have different amplitudes.

Example 2:
In fig.14.22 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.22

From the graph, we get two information:
(i) Information about period T
• Red and green do not reach maximums at the same instants.
• Red and green do not reach minimums at the same instants
• Red and green do not reach zero at the same instants
• So red and green do not have the same phase $\small{\left(\omega t \,+\,\phi \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6}  \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{6} \,-\,\frac{\pi}{4} \right)}$

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.  

◼ Based on this example, we can write:
Two simple harmonic motions may have the same amplitude $\small{A}$. But they can have different  phases.

Example 3:
In fig.14.23 below, the red and green curves represent two independent simple harmonic motions.

Fig.14.23

From the graph, we get two information:
(i) Information about period T
• Consider the magenta vertical dashed line. Both red and green reach maximum at the time indicated by this dashed line.
• The "horizontal distance between y-axis and the magenta vertical dashed line" indicates a time duration. Within this time duration, green completes two cycles. Red completes only one cycle.
• So it is clear that, $\small{T_{\text{green}}}$ is half of $\small{T_{\text{red}}}$
• So red and green do not have the same angular frequency $\small{\left(\omega \right)}$
• For this example, following equations are used:
Red: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{4}  \right)}$
Green: $\small{x(t)~=~4 \cos\left(\frac{\pi t}{2} \,-\,\frac{\pi}{4} \right)}$

(ii) Information about amplitude A
• The two yellow horizontal dashed lines indicate that, the amplitude of both red and green is 4 units.  

◼ Based on this example, we can write:
Two simple harmonic motions may have the same amplitude $\small{A}$. But they can have different angular frequencies.

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Now we will see a solved example:

Solved example 14.4
Which of the following function of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic ? Give the period for each case
(i) $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
(ii) $\small{\sin^2 \left(\omega t \right)}$
Solution:
Part (i):
1. We are given: $\small{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin\left(\omega t \right)~-~\cos\left(\omega t \right)}    & {~=~}    &{\sin\left(\omega t \right)~-~\sin\left(\frac{\pi}{2} - \omega t \right)}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2 \cos\left(\frac{\pi}{4} \right)\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2 \left(\frac{1}{\sqrt 2} \right)\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
\end{array}}$

◼ Remarks:

• 1 (magenta color):
Here we use the identity 6 in the list of trigonometric identities.
• 2 (magenta color):
Here we use the identity 20(d) in the list of trigonometric identities.

2. Based on this standard form, we can write:
The given expression represents a SHM with,
    ♦ Amplitude $\small{\sqrt 2}$ units
    ♦ Angular frequency $\small{\omega}$
    ♦ Phase constant $\small{- \frac{\pi}{4}}$

3. The standard form obtained in (1) can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t - \frac{\pi}{4} \right)}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left[2\pi~+~\left(\omega t - \frac{\pi}{4} \right) \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\sqrt 2\,\sin\left(\omega t + \frac{7\pi}{4} \right)}    \\
\end{array}}$

• So for the given SHM, the phase constant $\small{- \frac{\pi}{4}}$ is same as $\small{\frac{7\pi}{4}}$.

Part (ii):
1. We are given: $\small{\sin^2 \left(\omega t \right)}$
• Let us try to rearrange the expression into the standard form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin^2 \left(\omega t \right)}    & {~=~}    &{\frac{1\,-\,\cos\left(2\omega t \right)}{2}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{1}{2}~-~\frac{\cos\left(2\omega t \right)}{2}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{1}{2}~-~\frac{1}{2}\,\cos\left(2\omega t \right)}    \\
\end{array}}$

◼ Remarks:

• 1 (magenta color):
Here we use the identity 14 in the list of trigonometric identities.

2. In fig.14.24 below, two graphs are shown.

Fig.14.24

• Green represents: $\small{x(t)~=~0.5 \cos\left( 2\omega t \right)}$
    ♦ The two horizontal magenta dashed lines indicate that, the amplitude of green is 0.5 units.
    ♦ This graph is familiar to us. The equilibrium point is at O.

• Red represents: $\small{x(t)~=~0.5\,-\,0.5 \cos\left( 2\omega t \right)}$
    ♦ This is also a sinusoidal curve.
    ♦ The horizontal yellow dashed line and the x-axis indicate that, the distance between extreme points is 1 unit.
    ♦ So the amplitude of red is 1/2, which is 0.5 units.
    ♦ The upper horizontal magenta dashed line divides the red into two equal upper and lower parts. This dashed line passes through 0.5 on the y-axis. That means, equilibrium point of red is 0.5 units away from the origin.

3. Comparing the green with the standard form, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2\omega}    & {~=~}    &{\frac{2\pi}{T}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\omega}    & {~=~}    &{\frac{\pi}{T}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{T}    & {~=~}    &{\frac{\pi}{\omega}}    \\
\end{array}}$

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In the next section, we will see velocity and acceleration in simple harmonic motion.

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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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Chapter 4.15 - The Angular speed

In the previous section we saw the basic details of uniform circular motion and centripetal acceleration. In this section, we will see Angular speed.

We will write the steps:
1. Consider the circular motion of the object that we saw in the previous section. For convenience, it is shown again in fig.4.38 below:

Fig.4.38

• In a time interval of Δt, the object travels from P to P’. 
• During this time interval, the line OP turns through an angle of Δθ. 
    ♦ This Δθ  is called angular distance. 
2. From this 'angular distance', we can find 'the angle which the line OP turns in an interval of 1 s'.
• We simply divide the 'angular distance' by 'time required to cover that angular distance'. We get: 
• The angle which the line OP turns in 1 s = $\mathbf\small{\frac{\Delta \theta }{\Delta t}}$ 
3. The 'angle which the line (line which joins the object to the center) turns in  1 s' is called angular speed. It is denoted by $\mathbf\small{\omega}$ (Greek small letter omega) 
■ So we can write Eq.4.28: $\mathbf\small{\omega =\frac{\Delta \theta }{\Delta t}}$
4. The straight line between P and P’ is the 'displacement during Δt'
    ♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
    ♦ It is a vector quantity
• But the arc length PP’ is the 'distance traveled during Δt'
    ♦ It is denoted as $\mathbf\small{{\Delta s}}$ 
    ♦ It is not a vector quantity
5. We know that 'distance traveled' divided by time gives speed
• Speed is the magnitude of the velocity vector
    ♦ For circular motion, we denoted it as $\mathbf\small{|\vec{v}|}$
■ So we get: $\mathbf\small{|\vec{v}|=\frac{\Delta s}{\Delta t}}$
6. Now, using properties of arc, we have the relation: $\mathbf\small{angle=\frac{arc}{radius}}$
    ♦ Where angle is measured in radians
• So we get: arc length PP’ = Δs = Δθ×R
    ♦ Where R is the radius of the path
7. So we can put (Δθ×R) instead of Δs in (5). we get:
• Speed $\mathbf\small{|\vec{v}|=\frac{\Delta \theta \times R }{\Delta t}=R\frac{\Delta \theta}{\Delta t}}$
• But from (3), we have: $\mathbf\small{\frac{\Delta \theta }{\Delta t}=\omega}$
■ So we get Eq.4.29: speed $\mathbf\small{|\vec{v}|=R\, \omega}$
8. The time required for making one complete revolution around the circular path is called time period. It is denoted by T
An example:
(i) Let an object require 0.25 s to complete one revolution. Then it’s time period T = 0.25 s
(ii) This object requires only a ‘fraction of a second’ (¼ to be exact) to make one complete revolution
(iii) Obviously, It will make 4 complete revolutions in 1 s. 
• We obtain this result by dividing ‘1’ by ‘T’.
• That is., we take the reciprocal of T
9. The number of revolutions in one second is called frequency. It is denoted by ‘n’
■ So we can write Eq.4.30: Frequency $\mathbf\small{n=\frac{1}{T}}$
10. While calculating 'n', the 'T' goes to the denominator.
(i) In the above example, T (which is 0.25 s) is less than 1. 
• So when we calculate n, since T is in the denominator, we get a value ‘4’ which is greater than 1
(ii) What if T is greater than 1?
• That is., the object may take more than 1 s to complete one revolution
• In that case, since T is in the denominator, the ‘n’ will be less than ‘1’
■ That means, the object will not complete even one revolution in 1 s
11. Relations between $\mathbf\small{|\vec{v}|}$, T and n 
A. Relation between speed $\mathbf\small{|\vec{v}|}$ and T
(i) During an interval of ‘T’ s the object will travel a distance of 2πR along the circumference
(ii) We have the familiar relation for uniform speed: Distance = speed × time
(iii) So we get: 2πR = $\mathbf\small{|\vec{v}|}$ × T
Thus we get Eq.4.31: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{T}}$
B. Relation between speed $\mathbf\small{|\vec{v}|}$ and n
(i) From Eq.4.30, we have: $\mathbf\small{n=\frac{1}{T}}$
• From this, we get: $\mathbf\small{T=\frac{1}{n}}$
(ii) So we can use $\mathbf\small{\frac{1}{T}}$ instead of T in Eq.4.31
• We can write: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{\frac{1}{n}}}$
• So we get Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
12. Finally, we will see the relation between centripetal acceleration $\mathbf\small{|\vec{a_c}|}$ and $\mathbf\small{\omega}$
(i) In the previous section, we saw Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$  
(ii) In this section we derived Eq.4.29: $\mathbf\small{|\vec{v}|=R\, \omega}$
(iii) So we can use 'Rω' instead of $\mathbf\small{|\vec{v}|}$ in Eq.4.27
• We get: $\mathbf\small{|\vec{a_c}|=\frac{R^2 \omega ^2}{R}}$
• Thus we get Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$

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Now we will see an interesting case. We will write it in steps:
1. Consider a rotating circular disk of radius R cm shown in fig.4.39(a) below:

Fig.4.39

• It is rotating at a constant angular speed
2. Mark a point P on it's circumference
    ♦ Join P to the center O by the line OP. Then OP = R cm
• Mark another point Q on OP
    ♦ Let OQ = R₁ cm
3. Let during an interval of Δt s, the line OP turn through an angle of Δθ.
• Then, in that interval Δt, OQ will also turn through the same angle Δθ
• This is shown in fig.b
4. Since both P and Q turn through the same angle Δθ, we can write:
• Angular speed of Q = Angular speed of P
• Let this angular speed be $\mathbf\small{\omega}$
■ We can write: Angular speed of Q = Angular speed of P = $\mathbf\small{\omega}$
5. Now we will see linear speed
• Let the linear speed of P be $\mathbf\small{|\vec{v_P}|}$   
• Let the linear speed of Q be $\mathbf\small{|\vec{v_Q}|}$
6. During a time interval of Δt, P travels a distance of arc length PP'.
• During the same time interval of Δt, Q travels a distance of arc length QQ'.
7. But (arc length QQ') is less than (arc length PP')
• That is., even though the time interval is the same, the distances are not the same
■ Q travels a lesser distance in Δt
■ P travels a greater distance in Δt
■ Obviously, $\mathbf\small{|\vec{v_P}|}$ must be greater than $\mathbf\small{|\vec{v_Q}|}$
8. This can be proved in another way also:
(i) We have Eq.4.29 which gives the relation between linear speed and angular speed: $\mathbf\small{|\vec{v}|=R\, \omega}$
(ii) The angular speed $\mathbf\small{\omega}$ is constant for all points on the disc  
So we can write:
• $\mathbf\small{|\vec{v_P}|=R\, \omega}$ 
• $\mathbf\small{|\vec{v_Q}|=R_1 \, \omega}$ 
(iii) R₁ is lesser than R
■ So $\mathbf\small{|\vec{v_Q}|}$ will be lesser than $\mathbf\small{|\vec{v_P}|}$
■ When distance from the center decreases, the linear speed decreases

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Now we will see some solved examples
Solved example 4.13
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ? 
Solution:
• Number of revolutions in 100 s = 7
• So no. of revolutions in 1 s = n = 7/100 = 0.07
Part (a):
1. First we find linear speed $\mathbf\small{|\vec{v}|}$
• We can use Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}|=2\pi \times 12 \times 0.07}$ = 5.28 cm s^-1.
2. Once we find the linear velocity, we can use Eq.4.29 to find angular speed $\mathbf\small{\omega}$
• $\mathbf\small{|\vec{v}|=R\, \omega}$
• Substituting the values, we get: $\mathbf\small{5.28=12 \times \omega}$
• So $\mathbf\small{\omega}$ = 5.28/12 = 0.44 rad s^-1.  
Part (b):
1. The acceleration vector is always directed towards the center
• For each position of the insect, the acceleration is directed from that position to the center
■ That means, acceleration vector changes continuously. It is not a constant vector
2. But the magnitude of the acceleration vector is a constant
• To find that magnitude, we can use Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$ 
• Substituting the values, we get $\mathbf\small{|\vec{a_c}|= 0.44 ^2 \times 12}$ = 2.32 ms^-2. 

Solved example 4.14
A particle is in uniform circular motion along a circle whose center is at the origin in the xy plane. When the particle is at (4,0), it's velocity is $\mathbf\small{-5\hat{j}}$ ms^-1. Determine the velocity and acceleration when the particle is at P(0,-4) and Q(-2.83, 2.83)
Solution:
1. The velocity at (4,0) is given as $\mathbf\small{-5\hat{j}}$ ms^-1.
• The unit vector $\mathbf\small{\hat{j}}$ indicates that it is along the y axis
• The negative sign indicates that, it is towards the 'negative side of y axis'    
This situation is shown in fig.4.40(a) below:

Fig.4.40

2. The particle is said to be in uniform motion. So at any point along the path, $\mathbf\small{|\vec{v}|}$ will be 5 ms^-1.
• We just need to find the directions.
3. Also given that, the center is at the origin O and (4,0) is a point on the path
• So obviously, the radius R = 4 m
4. Part (a):
(i) P(0,-4) is on the y axis.
(ii) The particle is moving in the anti-clockwise direction
• So $\mathbf\small{\vec{v}}$ at P(0,-4) is $\mathbf\small{-5\hat{i}}$ ms^-1.    
(iii) We are also asked to find the acceleration
• In uniform circular motion, there is no linear acceleration. There is only centripetal acceleration $\mathbf\small{\vec{a_c}}$
• We can use Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$
• Substituting the values, we get: $\mathbf\small{|\vec{a_c}|}$ = 6.25 ms^-2.
• The direction of this acceleration is towards the center O
5. Part (b):
(i) Q(-2.83, 2.83) is neither on the x axis nor on the y axis
• First let us confirm that it is on the given circle.
(ii) For that, we find it's distance from the origin.
• We have: Distance from origin = $\mathbf\small{\sqrt{(-2.83)^2+2.83^2}}$ = 4 m
• So the point is indeed on the circle
(iii) Consider the following points:
x cordinate and y ccordinate of Q are equal in magnitudes
x coordinate is negative and y cordinate is positive
So obviously, Q is on a line inclined at 45^o to the x axis. And it is in the second quadrant.  This is shown in fig.b
• Let the velocity vector at Q be $\mathbf\small{\vec{v'}}$
• This $\mathbf\small{\vec{v'}}$ will be perpendicular to OQ
• So $\mathbf\small{\vec{v'}}$ makes an angle 45^o with the x axis 
(iv) Thus we get: 
• Magnitude of the x component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × cos 45 = 5 × cos 45 = 3.536 ms^-1.
• Magnitude of the y component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × sin 45 = 5 × sin 45 = 3.536 ms^-1. 
• So $\mathbf\small{\vec{v'}=3.536\, \hat{i}+3.536\, \hat{j}}$
(v) Direction of $\mathbf\small{\vec{v'}}$ is obtained as follows:
• $\mathbf\small{\tan \theta =\frac{|\vec{v'_y}|}{|\vec{v'_x}|}=\frac{3.536}{3.536}=1}$
• So θ = tan^-11 = 45° 
• We can write: $\mathbf\small{\vec{v'}}$ makes an angle of 45° with the x axis

Solved example 4.15
If the angular speeds of the minute hand and hour hand of a watch are $\mathbf\small{\omega_m}$ and  $\mathbf\small{\omega_h}$ respectively, find the ratio $\mathbf\small{\frac{\omega _m}{\omega _h}}$
Solution:
■ Angular speed is the 'angle turned in 1 s'
1. The angular speed of a minute hand:
(i)  The minute hand makes one complete revolution in 60 minutes. 
    ♦ 60 minutes is (60 × 60) = 3600 seconds
(ii) In other words, the minute hand turns through 2π radians in 3600 s
(iii) So it turns $\mathbf\small{\frac{2\pi }{3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _m=\frac{2\pi }{3600}}$
2. The angular speed of a hour hand:
(i)  The hour hand makes one complete revolution in 12 hours. 
    ♦ 12 hours is (12 × 60 × 60) = (12 ×3600) seconds
(ii) In other words, the hour hand turns through 2π radians in (12 × 3600) s
(iii) So it turns $\mathbf\small{\frac{2\pi }{12 \times 3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _h=\frac{2\pi }{12 \times 3600}}$
3. So we get: $\mathbf\small{\frac{\omega _m}{\omega _h}=\frac{2\pi }{3600}\div \frac{2\pi }{12 \times 3600}=\frac{2\pi }{3600}\times \frac{12 \times 3600 }{2 \pi} }$ = 12
■ Thus we get: angular speed of the minute hand is 12 times that of the hour hand

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In the next section, we will see Relative velocity in two dimensional motion.

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