More Examples of SHM

Examples of SHM

Motion of a Loaded Spring

Consider a spring having spring constant k suspended from a rigid support at one end and a certain load \(mg\) is attached at its free end. Let, \(l\) be the extension produced on the spring due to the load, \(F=mg\). Thus, within elastic limit, from Hooke's law, \begin{align*} F\propto l \end{align*}\begin{align*} \therefore F=kl \end{align*} Again, let us produce certain extra extension y by appplying an extra force. If \(F'\) be the total force applied, then,\begin{align*} F'\propto l \end{align*}\begin{align*} \therefore F'=k(l+y) \end{align*} Then, the extra force = \(F'-F=ky\). Thus the restoring force acting on the spring is \(-ky\). From Newton's second law, \begin{align*} ma=-ky \end{align*}\begin{align*} \therefore a=-\frac{k}{m}y \end{align*} This is the required equation of motion of a loaded spring. This equation shows that the motion of the loaded spring is simple harmonic. Comaparing it with the standard equation of SHM, we get, \begin{align*} \omega =\sqrt{\frac{k}{m}} \end{align*} And, the time period of oscillation is, \begin{align*} T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}} \end{align*}

Torsional pendulum

It is a heavy metallic disc suspended vertically from a rigid support by a string such that the disc can oscillate about a vertical axis in a horizontal plane.

Let, the disc be rotated in a horizontal plane by an angle \(\theta\). Then the string also gets twisted by the same twisting angle. Because of modulus of rigidity of the string, the disc experiences a restoring torque given by \(\tau=-c\theta\) where c is the restoring torque per unit twisting angle or torsional rigidity.

If I be the moment of inertia of the disc about the vertical axis passing through its and \(\alpha\) be the angular acceleration, then the deforming torque is given by \(I\alpha\). Within elastic limit, \begin{align*} I\alpha=-c\theta \end{align*}\begin{align*} or, \alpha=-\frac{c\theta}{I} \end{align*} This is the required equation of motion of the torsional pendulum which shows that the acceleration is directly proportional to the displacement from mean position. Comparing it with the standard equation of SHM, we get,\begin{align*} \omega =\sqrt{\frac{c}{I}} \end{align*} And, the time period of oscillation is, \begin{align*} T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{c}} \end{align*}

Helmholtz Resonator

A device used to find out what particular frequency are present in a given note is called a resonator. The resonators of different natural frequencies are used to detect the frequencies comprising the given note. It consists of large spherical or cylindrical glass or metal vessel containing air with a narrow neck N through which it receives the complex note to be analysed from outside air and a narrow aperture O at the opposite end to be plugged into the ear. The air in the neck serves as the air piston and performs an oscillatory motion like a piston in a cylinder containing air or a mass suspended from a spring.

Let, \(l\) be the length of the neck and \(\alpha\) be the cross section of the neck. Then, mass of air piston would be \(l\alpha\rho\) where \(\rho\) is the density of air inside the resonator. If this air piston be forced inward through a small distance x, the decrease in volume of air inside the vessel is \(\partial V=-\alpha x\) which causes the slight increase in pressure \(\partial P\). If k be the bulk modulus of elasticity of the air inside the vessel and V be the volume of the vessel, then,\begin{align*} k=\frac{\partial P}{\frac{\partial V}{V}} \end{align*}\begin{align*} \partial P=-k\frac{\partial V}{V} \end{align*} Thus, the force acting on the air piston outward is, \(\partial P\alpha=-k\frac{\partial V}{V}\alpha\). If \(\frac{\mathrm{d^2x} }{\mathrm{d} t^2}\) be the acceleration of the piston, the force acting on it is \(l\alpha\rho\frac{\mathrm{d^2x} }{\mathrm{d} t^2}\). Thus, we can write, \begin{align*} l\alpha\rho\frac{\mathrm{d^2x} }{\mathrm{d} t^2}=-k\frac{\partial V}{V}\alpha \end{align*}\begin{align*} or,l\alpha\rho\frac{\mathrm{d^2x} }{\mathrm{d} t^2}=\frac{-k\alpha^2x}{V} \end{align*}\begin{align*} or,\frac{\mathrm{d^2x} }{\mathrm{d} t^2}=\frac{-k\alpha}{Vl\rho} \end{align*}\begin{align*} \therefore\frac{\mathrm{d^2x} }{\mathrm{d} t^2}=-\mu x \end{align*} Thus, the air piston executes SHM with time period, \begin{align*} T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{Vl\rho}{k\alpha}} \end{align*} We know, velocity of sound in air is \(v=\sqrt{\frac{k}{\rho}}\). So, the time period can also be written as, \begin{align*} T=\frac{2\pi }{\omega }=\frac{2\pi}{v}\sqrt{\frac{Vl}{\alpha}} \end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.