Angular Simple Harmonic Motion

Angular Simple Harmonic Motion (ASHM) is a fascinating concept in physics that extends the principles of linear simple harmonic motion to rotational systems. Unlike linear motion, where objects oscillate back and forth along a straight line, ASHM involves oscillations about a fixed axis, like the swinging of a pendulum or the rotation of a balance wheel in a clock. This type of motion is characterized by a restoring torque proportional to the angular displacement, leading to periodic motion. In real life, ASHM is seen in the oscillations of a playground swing, the rotation of fan blades, and even in the periodic motion of molecules in certain chemical structures. Understanding ASHM helps us design efficient mechanical systems, predict natural phenomena, and enhance technologies that rely on precise rotational movements.

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1. Angular Simple Harmonic Motion
2. Solved Examples Based on Angular Simple Harmonic Motion
3. Summary

Angular Simple Harmonic Motion

Angular Simple Harmonic Motion (ASHM) is a type of periodic motion where an object oscillates around a fixed axis under the influence of a restoring torque that is directly proportional to the angular displacement. Unlike linear simple harmonic motion, which involves movement along a straight line, ASHM describes rotational motion, like the swinging of a pendulum or the torsional oscillations of a rod.

The general equation of linear SHM is given by $x=A\sin (\omega t+\alpha )$
Similarly, The general equation of angular SHM is given by $\theta ={\theta }_0\sin (\omega t+\varphi )$
where $\theta$ and ${\theta }_0$ are the angular displacement and angular amplitude of the bob respectively, as shown in the below figure

If $\mathrm{l}=$ length of the bob then we can write $\theta =\frac{x}{l}$ and ${\theta }_0=\frac{A}{l}$.
Similarly, The angular velocity if the bob which is in angular SHM is given by
$$\begin{array}{rl}& \dot{\theta }=\frac{d\theta }{dt}={\theta }_0\omega \mathrm{Cos}(\omega t+\varphi )\\ & \text{ or }\dot{\theta }=\omega \sqrt{{\theta }_0^2-{\theta }^2}\end{array}$$

Similarly, The angular acceleration if the bob which is in angular SHM is given by
$$\alpha =\frac{d^2\theta }{dt}=-{\theta }_0{\omega }^2\mathrm{Sin}(\omega t+\varphi )$$
or $\alpha =-{\omega }^2\theta$
And Thus restoring torque on the body is given as
$${\tau }_R=-I\alpha =-I{\omega }^2\theta$$

Thus we can state that in angular SHM, the angular acceleration of the body and the restoring torque on the body are directly proportional to the angular displacement of the body from its mean position and are directed toward the mean position.

Similarly, a basic differential equation for angular SHM can be written as

$\frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0$

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Solved Examples Based on Angular Simple Harmonic Motion

Example 1: A particle performing angular SHM has amplitude as $\pi$. If at $t=\frac{T}{4}$ its angular displacement was $\frac{\sqrt{3}\pi }{2}$, then what is the phase constant of SHM?

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

Solution:

The general equation of angular SHM is given by $\theta ={\theta }_0\sin (\omega t+\varphi )$
From the question we have
$$\begin{array}{rl}& \frac{\sqrt{3}\pi }{2}=\pi \sin (\frac{2\pi }{T}\ast \frac{T}{4}+\varphi )\\ & \Rightarrow \frac{\sqrt{3}}{2}=\sin (\frac{\pi }{2}+\varphi )=\mathrm{Cos}(\varphi )\\ & \Rightarrow \varphi =\frac{\pi }{6}\end{array}$$

Hence, the answer is the option (2).

Example 2: A particle performing angular SHM has amplitude as $\pi$. What is its angular velocity when its angular displacement is $\frac{\sqrt{3}\pi }{2}$ in terms of T where T= Time period of SHM?

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

Solution:

The angular velocity of the bob which is in angular SHM is given by:

$$\begin{array}{rl}& \dot{\theta }=\frac{d\theta }{dt}={\theta }_0\omega \mathrm{Cos}(\omega t+\varphi )\\ & \text{ or }\dot{\theta }=\omega \sqrt{{\theta }_0^2-{\theta }^2}\end{array}$$

Given- ${\theta }_0=\pi ,\theta =\frac{\sqrt{3}\pi }{2}$
$$\dot{\theta }=\frac{2\pi }{T}\sqrt{{\pi }^2-{\left(\frac{\sqrt{3}\pi }{2}\right)}^2}=\frac{2\pi }{T}\ast \pi \ast \sqrt{1-\frac{3}{4}}=\frac{{\pi }^2}{T}$$

Hence, the answer is the option (3).

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

Solution:

Time Period of Torsional Pendulum Case

$$T=2\pi \sqrt{\frac{I}{K}}$$
wherein
$I=$ moment of inertia
$K=$ torsional constant

From Figure

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

Hence, the answer is the option (2).

Example 4: One end of a massless spring of spring constant k and natural length $l_0$ is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through a fixed end, then the elongation of the spring will be :

1) $\frac{k-m{\omega }^2{l}_0}{m{\omega }^2}$
2) $\frac{m{\omega }^2{l}_0}{\mathrm{k}+m{\omega }^2}$
3) $\frac{m{\omega }^2{l}_0}{k-m{\omega }^2}$
4) $\frac{k+m{\omega }^2{l}_0}{m{\omega }^2}$

Solution:

Let elongation in spring is $x$,
Using the centripetal force equation
$$\begin{array}{rl}& {\mathrm{F}}_{\text{spring }}=\mathrm{m}{\omega }^2(l_0+\mathrm{x})\\ & \mathrm{kx}=\mathrm{m}{\omega }^2(l_0+\mathrm{x})\\ & \mathrm{kx}=\mathrm{m}{\omega }^2{l}_0+\mathrm{m}{\omega }^2\mathrm{x}\\ & \mathrm{x}(\mathrm{k}-\mathrm{m}{\omega }^2)=\mathrm{m}{\omega }^2{l}_0\\ & \mathrm{x}=\frac{\mathrm{m}{\omega }^2{l}_0}{\mathrm{k}-\mathrm{m}{\omega }^2}\end{array}$$
Hence, the answer is the option (3).

Summary

Angular Simple Harmonic Motion (ASHM) describes the periodic oscillation of an object around a fixed axis due to a restoring torque proportional to the angular displacement. The equations governing ASHM are analogous to those in linear SHM, with key parameters like angular displacement, velocity, and acceleration playing crucial roles. Practical applications and problems related to ASHM, such as calculating phase constants, angular velocities, and frequencies of oscillation, are essential in understanding the behavior of rotational systems in both natural and engineered contexts.