Oscillation Of Pendulum

The oscillation of a pendulum is a fundamental concept in physics, where a mass (bob) suspended from a fixed point swings back and forth due to the force of gravity. This simple harmonic motion is not just a textbook phenomenon but is observed in various real-life applications. From the rhythmic ticking of a grandfather clock to the calming motion of a swing in a park, pendulums demonstrate the principles of periodic motion. Understanding these oscillations helps us grasp essential concepts in mechanics, such as energy conservation and resonance, which have broader implications in engineering, architecture, and even timekeeping.

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This Story also Contains

1. What is a Simple Pendulum?
2. The Time Period of Oscillation of a Simple Pendulum (T)
3. Various cases
4. Solved Examples Based on Oscillation of Pendulum
5. Summary

What is a Simple Pendulum?

A simple pendulum is a basic mechanical system consisting of a point mass, known as the bob, suspended from a fixed point by a lightweight, inextensible string or rod. When the bob is displaced from its equilibrium position and released, it swings back and forth under the influence of gravity, exhibiting periodic motion. The key assumptions for a simple pendulum are that the string or rod does not stretch, the bob is a point mass, and there is no air resistance or friction at the pivot point. The time period of its oscillation depends on the length of the string and the acceleration due to gravity, but it is independent of the mass of the bob and the amplitude of the swing, as long as the angles involved are small.

The Time Period of Oscillation of a Simple Pendulum (T)

The time period of oscillation T of a simple pendulum is the time it takes for the pendulum to complete one full back-and-forth swing. When the bob is displaced to position B, through a small angle from the vertical as shown in the below figure.

Then Bob will perform SHM and its time period is given as

$T=2 \pi \sqrt{\frac{l}{g}}$

where

m=mass of the bob

l = length of pendulum

g = acceleration due to gravity.

key points

1. The time period of a simple pendulum is independent of the mass of the bob. i.e If the solid bob is replaced by a hollow sphere of the same radius but different mass, the time period remains
unchanged.

2. $T \propto \sqrt{l}$ where l is the distance between the point of suspension and the centre of mass of the bob and is called effective length.

3. The period of a simple pendulum is independent of amplitude as long as its motion is simple harmonic.

Pendulum in a Lift

1. The time period of a simple pendulum, If the lift is at rest or moving downward /upward with constant velocity.

$T=2 \pi \sqrt{\frac{l}{g}}$

where

$\begin{aligned} & l=\text { the length of the pendulum } \\ & g=\text { acceleration due to gravity. }\end{aligned}$

2. The time period of a simple pendulum, If the lift is moving upward with constant acceleration a

$T=2 \pi \sqrt{\frac{l}{g+a}}$

where

$l=$ the length of the pendulum
$g=$ acceleration is due to gravity.
$a=$ acceleration of the pendulum.

3. The time period of a simple pendulum If the lift is moving downward with constant acceleration a

$T=2 \pi \sqrt{\frac{l}{g-a}}$

where

$l=$ the length of the pendulum
$g=$ acceleration is due to gravity.
$a=$ acceleration of the pendulum.

4. The time period of a simple pendulum, If the lift is moving downward with acceleration a = g

$
T=2 \pi \sqrt{\frac{l}{g-g}}=\infty
$

It means there will be no oscillation in a pendulum as here $g_{\text {eff }}=0$
Similarly in the case of a satellite and at the centre of the earth the $g_{e f f}=0$ so in these cases, effective acceleration becomes zero and the pendulum will stop.

5. The time period of a simple pendulum whose point of suspension moving horizontally with acceleration 'a'

For the above figure $g_{\text {eff }}=\left(g^2+a^2\right)^{\frac{1}{2}}$
$
T=2 \pi \sqrt{\frac{l}{\left(g^2+a^2\right)^{\frac{1}{2}}}}
$

Where
$l=$ the length of pendulum
$g=$ acceleration is due to gravity.
$a=$ acceleration of the pendulum.

6. The time period of a simple pendulum accelerating down an incline

In this case $g_{\text {eff }}=g \cos \theta$
$
T=2 \pi \sqrt{\frac{l}{g \cos \Theta}}
$
where
$l=$ the length of the pendulum
$g=$ acceleration is due to gravity.
$\theta=$ angle of inclination

The Time Period of the Pendulum in a Liquid

If we immerse a simple pendulum in a liquid, the bob of the pendulum will experience a buoyant force in an upward direction in addition to the other forces such as gravity and tension.

If bob a simple pendulum of density $\sigma$ is made to oscillate in some fluid of density $\rho$ (where $\rho<\sigma$ ).

Then the buoyant force is given as $F_B=V \rho g$
As buoyant force will oppose its weight therefore $F_{n e t}=m g_{e f f}=m g-F_B$

And for the above figure let bob is displaced for a small displacement x and is at an angle $\theta$ with the verticle.

For small displacement $x$ of the bob, restoring force
$
\begin{aligned}
& F_{\text {rest }}=(m g-V \rho g) \sin \theta=-(m g-V \rho g) \frac{x}{l} \\
& \text { and acceleration }=-\left(g-\frac{V \rho q}{m}\right) \frac{x}{l}
\end{aligned}
$

On comparing with the standard equation of SHM, $a=-\omega^2 x$, we get
$
\begin{aligned}
& \omega=\sqrt{\frac{\left(g-\frac{V_{\rho g}}{m}\right)}{l}}=\sqrt{\frac{g}{l}\left(1-\frac{\rho}{\sigma}\right)} \\
& \text { and } T=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}}
\end{aligned}
$

The Time Period of the Second's Pendulum

Second, Pendulum: It is that simple pendulum whose time period of vibrations is two seconds.

Putting T=2 sec in $T=2 \pi \sqrt{\frac{l}{g}}$ we get the Length of a second’s pendulum is nearly 1 meter on the earth's surface.

If the length of the pendulum is comparable to the radius of the earth

then $T=2 \pi \sqrt{\frac{1}{g\left(\frac{1}{l}+\frac{1}{R}\right)}}$

where

$\begin{aligned} & l=\text { length of pendulum } \\ & g=\text { acceleration due to gravity. } \\ & R=\text { Radius of earth }\end{aligned}$

Various cases

A. If $l \ll R$, then $\frac{1}{l} \gg \frac{1}{R} \quad$ so $\quad T=2 \pi \sqrt{\frac{l}{g}}$
B. If $l \gg R($ or $l \rightarrow \infty)$ then $\frac{1}{l}<\frac{1}{R}$
so $T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6.4 \times 10^6}{10}} \cong 84.6$ minutes
and it is the maximum time period which an oscillating simple pendulum can have
C. If $l=R \quad$ so $\quad T=2 \pi \sqrt{\frac{R}{2 g}} \cong 1$ hour

Solved Examples Based on Oscillation of Pendulum

Example 1: A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :

1) 92 ± 2 s

2) 92 ± 5.0 s

3) 92 ± 1.8 s

4) 92 ± 3 s

Solution:

Time period of oscillation of the simple pendulum

$
T=2 \pi \sqrt{\frac{l}{g}}
$
wherein
$\mathrm{I}=$ length of pendulum
$\mathrm{g}=$ acceleration due to gravity.
The sum of all observations $=90+91+95+92=368$
Average $=\frac{368}{4}=92$
The sum of modulus observations $=2+1+3=6$
$\therefore$ Average error $=\frac{6}{4}=1.5$, rounded off 2 .
$\therefore$ Final answer $=92 \pm 2$

Hence, the answer is the option (1).

Example 2: A pendulum clock loses 12 s a day if the temperature is 40⁰C and gains 4 s a day if the temperature is 20⁰C. The temperature at which the clock will show the correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively :

1) 25⁰C; α = 1.85×10⁻⁵/⁰C

2) 60⁰C; α =1.85×10⁻⁴/⁰C

3) 30⁰C; α=1.85×10⁻³/⁰C

4) 55⁰C; α = 1.85×10⁻²/⁰C

Solution:

Time period of oscillation of the simple pendulum

$
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& \text { wherein } \\
& \mathrm{I}=\text { length of pendulum } \\
& \mathrm{g}=\text { acceleration due to gravity. } \\
& \text { Times lost/gained per day }=\frac{1}{2} \times \alpha \times \Delta \theta \times 86,400
\end{aligned}
$
wherein

Where $\alpha$-Co-efficient of linear expansion.
$
\begin{aligned}
& 12=\frac{1}{2} \times(40-\theta) \times 86400 \\
& 4=\frac{1}{2} \times(\theta-20) \times 86400
\end{aligned}
$

On dividing we get
$
\begin{aligned}
& 3=\frac{40-\theta}{\theta} \\
\Rightarrow & 3 \theta-60=40-\theta \\
\therefore & 4 \theta=100 \Rightarrow \theta=25^{\circ} \mathrm{C}
\end{aligned}
$

After putting the value of $\theta$ we get $\alpha$
$
\alpha=1.85 \times 10^{-5}
$

Hence, the answer is the option (1).

Example 3: A child swinging on a swing in a sitting position stands up, then the time period of the swing will :

1) increase

2) decrease

3) remains same

4) increases if the child is long and decreases if the child is short.

Solution:

Time period of oscillation of the simple pendulum

$\begin{aligned} & T=2 \pi \sqrt{\frac{l}{g}} \\ & \text { wherein } \\ & \mathrm{I}=\text { length of pendulum } \\ & \mathrm{g}=\text { acceleration due to gravity. }\end{aligned}$

The centre of gravity shifted upwards so the length from the point of suspension will decrease.

Hence, the answer is the option (2).

Example 4: The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t₀ in air. Neglecting the frictional force of water and given that the density of the bob is (4/3) x 1000 kg/m³. What relationship between t and t₀ is true?

1) t = t₀

2) t = t₀/2

3) t = 2t₀

4) t = 4t₀

Solution:

The time period of oscillation of a simple pendulum

$
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& \text { wherein } \\
& \mathrm{I}=\text { length of pendulum } \\
& \mathrm{g}=\text { acceleration due to gravity. } \\
& t_0=2 \pi \sqrt{l / g} \ldots \ldots . .(i)
\end{aligned}
$

Due to the upthrust of water on the top, its apparent weight decreases upthrust $=$ weight of the liquid displaced
$
\therefore \quad \text { Effective weight }=m g-(V \sigma g)=V \rho g-V \sigma g
$
$V \rho g^{\prime}=V g(\rho-\sigma)$, where $\sigma$ is density of water
or $g^{\prime}=g\left(\frac{\rho-\sigma}{\rho}\right)$
$
\begin{aligned}
& \therefore t=2 \pi \sqrt{l / g^{\prime}}=2 \pi \sqrt{\frac{l \rho}{g(\rho-\sigma)}} \ldots \ldots \ldots \text { (ii) } \\
& \therefore \quad \frac{t}{t_0}=\sqrt{\frac{l \rho}{g(\rho-\sigma)} \times \frac{g}{l}}=\sqrt{\frac{\rho}{\rho-\sigma}}=\sqrt{\frac{4 \times 1000 / 3}{\left(\frac{4000}{3}-1000\right)}}=2
\end{aligned}
$
or $t=t_0 \times 2=2 t_0$

Hence, the answer is the option (3).

Example 5: The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

1) remain unchanged

2) increase towards a saturation value

3) first increase and then decrease to the original value

4) First decrease and then increase to the original value

Solution:

Time period of oscillation of the simple pendulum

$T=2 \pi \sqrt{\frac{l}{g}}$

wherein

l = length of pendulum

g = acceleration due to gravity

For a pendulum, $T=2 \pi \sqrt{\frac{l}{g}}$ where l Is measured up to the centre of gravity. The centre of gravity of the system is at the centre of the sphere when the hole is plugged. When unplugged, water drains out. The Centre of gravity goes on descending. When the bob becomes empty, the centre of gravity is restored to the centre.

$\therefore$ The length of the pendulum first increases, then decreases to the original value.

$\therefore$T would first increase and then decrease to the original value.

Hence, the answer is the option (3).

Summary

The oscillation of a simple pendulum is a fundamental concept in physics, demonstrating periodic motion influenced by gravity. Its time period, independent of mass, depends on the length of the pendulum and gravitational acceleration. Variations in conditions, such as movement in a lift or immersion in a liquid, alter the effective gravitational force and thus the pendulum's time period, offering insights into broader mechanical principles and real-world applications.