Waves in Strings, Rods and Gases

Velocity of transverse waves along a stretched string

Consider a symmetrical pulse moving along a stretched string with speed \(v\). Consider, a small segment AB of the pulse of length \(\Delta l\) forming an arc of a circle of radius \(R\) and making an angle \(2 \theta\) at the centre \(O\) of the circle. A tension \(T\) in the string pulls this segment tangentially at each end. The horizontal components of \(T\) ie. \(T\cos\theta\) cancel each other but the vertical components \(T\sin\theta\) add to form radial restoring force \(F\) given by \(F=2T\sin\theta\). For small angles, it becomes \(F=2T\theta\) as \(\theta\) is nearly equal to \(\sin\theta\) form small angles. Also, we hav \(2\theta=\Delta l/R\). Therefore, \(F=T.\frac{\Delta l}{R}\). If \(u\) be the mass per unit length of the string then, the mass of the segment of length \(\Delta l\) is \(m=\mu \Delta l\). Since, this string segment is moving in an arc of a circle, the force defined acts as a centripetal force producing a centripetal acceleration towards the centre of circle given by, \(a=\frac{v^2}{R}\), where, \(v\) is the velocity of the wave. So, the centripetal force is given by \(F=\frac{mv^2}{R}\). Equating the above expressions for force, we get, \begin{align*} \frac{T\Delta l}{R}=\frac{mv^2}{R} \end{align*}\begin{align*} or, \frac{T\Delta}{R}=\mu \Delta l \end{align*}\begin{align*} \therefore v=\sqrt \frac {T}{\mu} \end{align*}This is the velocity of the pulse and any other wave on the same string under the same tension.

Modes of vibration in a stretched string

Since, the ends of a stretched string are fixed, these are the positions of nodes in the same wire. When the string is plucked at the middle, an antinode is formed.This is the simplest node of vibration. Distance between two consecutive node is \(L=\lambda /2 and \lambda=2L\) where, \(\lambda\) is the measuremnt of transverse wave in the string and L is the length of the stiring. The frequency of vibration is given by, \(f_{o}=\frac{v}{\lambda}=\frac{v}{2L}\), where, \(v\) is the velocity of the transverse wave. This is the fundamental frequency or frequency of the first harmonic. It is the lowest frequency produced by vibrating string.
If the string is plucked at a point one quarter of its length from one end, the string vibrates in two segments. This mode of vibration is called the first overtone. The vibration can also be set when the vibrating string in fundamental mode is lightly touched at the middle. So, three nodes and two antinodes are formed in the string as shown. If \(\lambda\) be the wavelength and \(f_{1}\) be the frequency of the resulting stationary wave, then \(L=\lambda\) and frequency of the wave \(f_{1}=\frac{v}{\Lambda}=\frac{v}{L}=2_f{o}\). Thus, the frequency of the first overtone is two times the fundamental frequency. This is called second harmonics.
If the string is made to vibrate in three segment by touching it(when vibrating in fundamental mode) at one-third of its
length from one end, one additional node and the additional antinode are produced. If \(\lambda\) be the wavelength and \(f_{2}\) be the frequency of the wave, then, \(L=\frac{3\lambda}{2}\) and \(\frac{2L}{3}\). The frequency is then \(f=\frac{v}{\lambda}=\frac{3v}{2L}=3f_{o}\). Thus, the frequency of second overtone is three times the fundamental frequency. This is also called third harmonics.
Hence, it is observed that all harmonics can be formed in stretched string and the ratio of the frequencies are \(f_{o}:f_{1}:f_{2}:f_{3}...=1:2:3:4...\)

Waves in rods

The modes of vibration in a metal rod depend on the position at which the rod is clamped. When the rod is clamped tightly at middle, and is struck along the length at a free end, stationary longitudinal waves are formed on the rod. Since the ends are free to vibrate, displacement antinodes are formed at the ends of the rod and and a dislacement node at the middle as all the atoms are at rest there. In the simplest mode of vibration of rod, \(L=\lambda/2\). If \(v\) is the velocity of the wave, the frequency is given by, \(f=\frac{v}{\lambda}=\frac{v}{2L}\) and so on for the higher modes.

Velocity of longitudinal waves in gases

At \(t=0\). the piston is pushed in suddenly with velocity \(v\), so the pressure increases and becomes \(P+\Delta P\). This will create a disturbance which travels with velocity say \(v_{w}\). In time \(t\) the disturbance travels a distance \(v_{w}t\) and reaches to \(C\). But at the same time, the piston moves a distance \(vt\) and reaches \(B\). The original volume of the gas that is set in motion in time \(t\) is \(v_{w}t\times A\) where, \(A\) is the cross section of the tube. Therefore, the mass of the gas under motion is \(v_{w}tA\rho\). Since the gas is moving with velocity \(v\), the change in linear momentum of the gas is \(v_{w}tA\rho V\). Due to the inward movement of the piston, let \(\Delta V\) be the compression in the volume of gas between A and C. Therefore, \(\Delta V\) is volume of \(AB\) which is equal to \(Avt\). So, for increase in pressure \(\Delta P\), the bulk modulus of elasticity is,\begin{align*} B=\frac{\Delta P}{\frac{\Delta V}{V}}=\frac{\Delta P}{\frac{Avt}{Av_{w}t}}=\frac{\Delta P}{\frac{v}{v_{w}}}\end{align*}\begin{align*} \therefore \Delta P=B\frac{v}{v_{w}} \end{align*}Therefore, impulse on the moving gas in time \(t\) is \(\Delta PAt=B\frac{v}{v_{w}}At\). Applying impulse momentum theorem,\begin{align*}B\frac{v}{v_{w}}At=(v_{w}tA\rho)v \end{align*}\begin{align*} v_{w}=\sqrt{\frac{B}{\rho}}\end{align*} This gives the velocity of longitudinal wave in gas.

Velocity of sound in gas

Newton's formula

Newton assumed that when sound waves travel through a gaseous medium, the variation in temperature produced by the compression and rarefaction are negligible. It is because the process is so slow that heat exchange with the surrounding takes place. The condition therefore is isothermal and Boyle's law can be applied as,\begin{align*} PV=constant \end{align*} Differentiaitng we get,\begin{align*}PdV+VdP=0 \end{align*}\begin{align*} p=\frac{dP}{-\frac{dV}{V}}=B \end{align*} The negative sign indicates the pressure volume relationship. Therefore, the velocity of sound is,\begin{align*} v=\sqrt{\frac{B}{\rho}}=\sqrt{\frac{P}{\rho}} \end{align*} Substituting \(P=\rho gh=1.01\times 10^5 N/m^2\) and \(\rho=1.29 kg/m^3\) for air at STP, we get, \(v=280 m/s\). But at STP, velocit of sound through air is \(332m/s\). This inaccurate value is obtained de to some discrepancies in Newton's assumption.

Laplace's correction over Newton's formula

The process of compression or rarefaction is so rapid that neither heat is transferred to the surrounding during compression nor heat is taken from the surrounding during rarefaction. So, the temperature in different regions does not remain constant. Thus, when sound wave propagates through the gas, the condition is adiabatic process. So, for adiabatic process,\begin{align*} PV^\gamma=constant---1 \end{align*} where, \(\gamma=C_{p}/C_{v}\). Differentiating 1 and on solving, we get, \(\gamma P=B\). Hence, velocity of sound is,\begin{align*} v=\sqrt{\frac{\gamma P}{\rho}}\end{align*} For air, \(\gamma=1.4\). So at STP, we get, \(v=331 m/s\). This result is in good agreement with the experimental value at STP.

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.