Time Period Of Torsional Pendulum

Time Period of Torsional Pendulum

The time period of a torsional pendulum refers to the time it takes for the pendulum to complete one full oscillation, rotating back and forth about its equilibrium position. This time period is determined by the moment of inertia of the object and the torsional constant of the wire or rod from which it is suspended. Below is the figure of the Torsional pendulum which consists of a rigid object suspended by a wire attached at the top to a fixed end.

When the object is twisted through some angle $\theta$, the twisted wire exerts on the object a restoring torque and this restoring torque is proportional to the angular position.

That is $\tau=-k \theta$ where k is called the torsion constant of the support wire.

Applying Newton's second law for rotational motion, we find that

$\tau=-k \theta=I \frac{d^2 \theta}{d t^2} \Rightarrow \frac{d^2 \theta}{d t^2}=-\frac{k}{I} \theta$

So the Time Period of the Torsional pendulum is given as

$T=2 \pi \sqrt{\frac{I}{k}}$

where

I = moment of inertia

k = torsional constant

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Solved Examples Based on Time Period of Torsional Pendulum

Example 1: A pendulum bar magnet oscillates in the magnetic field with time period 'T'. If its mass is increased by four times then, its time period will be

1) 4 T
2) 2 T
3) T
4) $\frac{T}{2}$

Solution

Time Period of Torsional Pendulum Case

$
T=2 \pi \sqrt{\frac{I}{K}}
$
wherein
$I=$ moment of inertia
$K=$ torsional constant
We know, $I \propto M$
By increasing mass by 4 times the moment of Inertia also $\uparrow$ by 4
$
I^{\prime}=4 I
$
$T \propto \sqrt{I}$
$
\begin{aligned}
\frac{T^{\prime}}{T} & =\sqrt{\frac{I^{\prime}}{I}} \\
T^{\prime} & =2 T
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: What is the frequency of a torsional pendulum in which the disc has a mass of 2 Kg and radius = 30 cm? The torsional stiffness of the wire through which the disc is hanged with the ceiling is 20 N-m. ( Answer will be in Hz)

1) 2.4

2) 2

3) 4

4) 4.8

Solution:

As,

$
T=2 \pi \sqrt{\frac{I}{k}}
$

So, the frequency is
$
\begin{aligned}
& f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{I}} \\
& f=\frac{1}{2 \pi} \sqrt{\frac{20}{I}} S o, I=\frac{1}{2} M \cdot R^2=\frac{1}{2} \times 2 \times(0.3)^2 I=0.09 \mathrm{Kg}-\mathrm{m}^2, f=\frac{1}{2 \pi} \cdot \sqrt{\frac{20}{0.09}}=2.4 \mathrm{~Hz}
\end{aligned}
$

Hence, the answer is the option (1).

Example 3: A ring is hung on a nail. It can oscillate without slipping or sliding (i) in its plane with a period $T_1$ and (ii) back and forth in a direction perpendicular to its plane with a period $T_2$ The ratio $\frac{T_1}{T_2}$ will be :

1) $\frac{\sqrt{2}}{3}$
2) $\frac{2}{3}$
3) $\frac{2}{\sqrt{3}}$
4) $\frac{3}{\sqrt{2}}$

Solution:

$\begin{aligned} & \frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}} \\ & I_1=M R^2+M R^2=2 M R^2 \\ & I_2=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2 \\ & \frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{2 * 2}{3}}=\frac{2}{\sqrt{3}}\end{aligned}$

Hence, the answer is the option (3).

Example 4: Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its centre 'O' and can rotate freely in the horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in the figure. The rod is gently pushed through a small angle and released. The frequency of the resulting oscillation is :

1) $\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
2) $\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}}$
3) $\frac{1}{2 \pi} \sqrt{\frac{3 k}{m}}$
4) $\frac{1}{2 \pi} \sqrt{\frac{2 k}{m}}$

Solution:

Time Period of Torsional Pendulum Case

$T=2 \pi \sqrt{\frac{I}{K}}$

wherein

I = moment of inertia

K = torsional constant

From Figure

$\begin{aligned} & x=\frac{l}{2} \sin \theta=\left(\frac{l}{2} \times \theta\right) \\ & \&^\iota=\overrightarrow{F_x} \cdot \vec{s} \\ & =-2 k x \times\left(\frac{l}{2}\right) \\ & \iota=-2 \times k \times \frac{l}{2} \Theta \times \frac{l}{2}=I \alpha=\frac{m l^2}{12} \alpha \\ & \Rightarrow \frac{K e^2}{2} \Theta=\frac{-m l^2}{12} \alpha \\ & \alpha=\frac{6 k}{m} \Theta=\text { so } w=\sqrt{\frac{6 k}{m}} \\ & f=\frac{w}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{m}}\end{aligned}$

Hence, the answer is the option (2).

Example 5: A rod of mass 'M' and length '2L' is suspended in its middle by a wire. It exhibits torsional oscillations. If two masses each of 'm' are attached at a distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to:(give answer till 3 decimal places)

1) 0.375

2) 0.175

3) 0.575

4) 0.775

Solution:

Time Period of Torsional Pendulum Case

$\begin{aligned} & T=2 \pi \sqrt{\frac{I}{k}} \\ & \text { - wherein } \\ & I=\text { moment of inertia } \\ & k=\text { torsional constant } \\ & \text { frequency } f=\frac{1}{2 \pi} \sqrt{\frac{k}{I}} \\ & f_1=\frac{1}{2 \pi} \sqrt{\frac{k}{\left(\frac{M(2 L)^2}{12}\right)}} \\ & f_2=\frac{1}{2 \pi} \sqrt{\frac{k}{\left(\frac{M(2 L)^2}{12}\right)+2 m\left(\frac{L}{2}\right)^2}} \\ & f_2=0.8 f_1 \\ & \frac{m}{M}=0.375\end{aligned}$

Hence, the answer is the option (1).

Summary

The period of a simple harmonic motion (SHM) torsional pendulum is the time taken for the pendulum to make a full oscillation about its midpoint or rest position. The time it takes to do this is dependent upon the moment of inertia of the pendulum as well as the torsional constant of the wire or rod from which it hangs. Angular displacement together with restoring torque is part of this motion just like linear displacement together with restoring force in a mass-spring system resulting in SHM.