Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

The document Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
All you need of JEE at this link: JEE

Q.171. A copper rod of length l = 50 cm is clamped at its midpoint. Find the number of natural longitudinal oscillations of the rod in the frequency range from 20 to 50 kHz. What are those frequencies equal to? 

Ans. Since the copper rod is clamped at mid point, it becomes a mode and the two free ends will be anitinodes. Thus the fundamental mode formed in the rod is as^shown in the Fig. (a).

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

in this caseIrodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

So,

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

where E = Young’s modules and p is the density of the copper

Similarly the second mode or the first overtone in the rod is as shown above in Fig. (b).

Here  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Putting the given values of E and p in the general equation

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence the sought number of frequencies between 20 to 50 k Hz equals 4.

 

Q.172. A string of mass rn, is fixed at both ends. The fundamental tone oscillations are excited with circular frequency ω and maximum displacement amplitude amax. Find:
 (a) the maximum kinetic energy of the string;
 (b) the mean kinetic energy of the string averaged over one oscillation period. 

Ans. Let two waves  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev superpose and as a result, we have a standing wave (the resultant wave ) in the string of the form  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev
According to the problem 2 a = am.

Hence the standing wave excited in the string is

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev         (1)

or,  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev        (2)

So the kinetic energy confined in the string element of length dx, is given by

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence the kinetic energy confined in the string corresponding to the fundamental tone

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Because, for the fundamental tone, length of the string  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Integrating we get, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence the sought maximum kinetic energy equals, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

because for Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(ii) Mean kinetic energy averaged over one oscillation period

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,        Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.173. A standing wave Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev = a sin kx•cos cot is maintained in a homogeneous rod with cross-sectional area S and density p. Find the total mechanical energy confined between the sections corresponding to the adjacent displacement nodes. 

Ans. We have a standing wave given by the equation

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

The kinetic energy confined in an element of length dx of the rod

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

So total kinetic energy confined into rod

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev(3)

The potential energy in the above rod element

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

so, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or,  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus the total potential energy stored in the rod Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

so, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

To find the potential energy stored in the rod element we may adopt an easier way. We know that the potential energy density confined in a rod under elastic force equals :

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence the total potential energy stored in the rod

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence the sought mechanical energy confined in the rod between the two adjacent nodes

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.174. A source of sonic oscillations with frequency v = 1000 Hz moves at right angles to the wall with a velocity u = 0.17 m/s. Two stationary receivers R1 and R2  are located on a straight line, coinciding with the trajectory of the source, in the following succession: R1-source-R2-wall. Which receiver registers the beatings and what is the beat frequency? The velocity of sound is equal to v = 340 m/s. 

Ans. Receiver R1 registers the beating, due to the sound waves reaching directly to it from source and the other due to the reflection from the wall.
Frequency of sound reaching directly from S to R1

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Now frequency reaching to R1 after reflection from wall

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus the sought beat frequency

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.175. A stationary observer receives sonic oscillations from two tuning forks one of which approaches, and the other recedes with the same velocity. As this takes place, the observer hears the beatings with frequency v = 2.0 Hz. Find the velocity of each tuning fork if -their oscillation frequency is vo = 680 Hz and the velocity of sound in air is v = 340 m/s. 

Ans. Let the velocity of tuning fork is u. Thus frequency reaching to the observer due to the tuning fork that approaches the observer

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Frequency reaching the observer due to the tunning fork that recedes from the observer

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 Hence the sought value of u, on sim plifying and noting that u > 0

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.176. A receiver and a source of sonic oscillations of frequency vo = 2000 Hz are located on the x axis. The source swings harmonically along that axis with a circular frequency ω and an amplitude a = 50 cm. At what value of ω will the frequency bandwidth registered by the stationary receiver be equal to Av = 200 Hz? The velocity of sound is equal to v = 340 m/s. 

Ans. Obviously the maximum, frequency will be heard when the source is moving with maximum velocity towards the receiver and minimum frequency will be heard when the source recedes with maximum velocity. As the source swing harmonically its maximum velocity equals ω to. Hence

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

So the frequency band width Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

On simplifying (and taking + sign as Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.177. A source of sonic oscillations with frequency v= 1700 Hz and a receiver are located at the same point. At the moment t = 0 the source starts receding from the receiver with constant acceleration w = 10.0 m/s2. Assuming the velocity of sound to be equal to v = 340 m/s, find the oscillation frequency registered by the stationary receiver t = 10.0 s after the start of motion. 

Ans. It should be noted that the frequency emitted by the source at time t could not be received at the same moment by the receiver, becouse till that time the source will cover the distance Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev and the sound wave will take the further time Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRevto reach the receiver. Therefore the frequency noted by the receiver at time t should be emitted by the source at the time Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev Therefore

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev(1)

and the frequency noted by the receiver

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev(2)

Solving Eqns (1) and (2), we get

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.178. A source of sound with natural frequency v= 1.8 kHz moves uniformly along a straight line separated from a stationary observer by a distance l = 250 m. The velocity of the source is equal to η = 0.80 fraction of the velocity of sound. Find:
 (a) the frequency of sound received by the observer at the moment when the source gets closest to him;
 (b) the distance between the source and the observer at the moment when the observer receives a frequency v = vo

Ans. (a) When the observer receives the sound, the source is closest to him. It means, that frequency is emitted by the source sometimes before (Fig.) Figure shows that the source approaches the stationary observer with

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Hence from Eqns. (1) and (2) the sought frequency

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(b) When the source is right in front of O, the sound emitted by it will not be Doppler shifted because 0 = 90°. This sound will be received at O at time Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRevafter the source has v passed it. The source will by then have moved ahead by a distance Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev. The distance between the source and the observer at this time will be Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.179. A stationary source sends forth monochromatic sound. A wall approaches it with velocity u = 33 cm/s. The propagation velocity of sound in the medium is v = 330 m/s. In what way and how much, in per cent, does the wavelength of sound change on reflection from the wall? 

Ans.  Frequency of sound when it reaches the wall 

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

wall will reflect the sound with same frequency v'. Thus frequency noticed by a stationary observer after reflection from wall

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev since wall behaves as a source o f frequency v'

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev.

Hence the sought percentage change in wavelength 

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.180. A source of sonic oscillations with frequency vo = 1700 Hz and a receiver are located on the same normal to a wall. Both the source and the receiver are stationary, and the wall recedes from the source with velocity u = 6.0 cm/s. Find the beat frequency registered by the receiver. The velocity of sound is equal to'. v = 340 m/s. 

Ans. Frequency of sound reaching the wall.

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev(1)

Now for the observer the wall becomes a source of frequency v receding from it with velocity u Thus, the frequency reaching the observer

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev                  [Using (1)]

Hence the beat frequency registered by the receiver (observer)

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.181. Find the damping coefficient y of a sound wave if at distances r1 =  10 m and r2  = 20 m from a point isotropic source of sound the sound wave intensity values differ by a factor η = 4.5. 

Ans. Intensity of a spherical sound wave emitted from a point source in a homogeneous absorbing medium of wave damping coefficient y is given by

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

So, Intensity of sound at a distance r1 from the source

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

and intensity of sound at a distance r2 from the source

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

But according to the problemIrodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.182. A plane sound wave propagates along the x axis. The damping coefficient of the wave is γ = 0.0230 m-1. At the point x = 0 the loudness level is L = 60 dB. Find:
 (a) the loudness level at a point with coordinate x = 50 m;
 (b) the coordinate x of the point at which the sound is not heard any more. 

Ans.  (a)Loudness level in bells Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev (I0 is the theshold o f audibility.)

So, loudness level in decibells, Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus loudness level atIrodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Similarly  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(b) The point at which the sound is not heard any more, the loudness level should be zero.
Thus

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.183. At a distance r= 20.0 m from a point isotropic source of sound the loudness level L= 30.0 dB. Neglecting the damping of the sound wave, find:

(a) the loudness level at a distance r = 10.0 m from the source;
 (b) the distance from the source at which the sound is not heard. 

Ans.   (a) As there is no damping, so

 Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(b) Let r be the sought distance at which the sound is not heard.

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus for r > 0. 63 km no sound will be heard.

 

Q.184. An observer A located at a distance ζ= 5,0 m from a ringing tuning fork notes the sound to fade away ζ= 19 s later than an observer B who is located at a distance ζB = 50 m from the tuning fork. Find the damping coefficient β of `Oscillations of the tuning fork. The sound velocity v = 340 m/s. 

Ans. We treat the fork as a point source. In the absence of damping the oscillation has the form

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Because of the damping of the fork the amplitude of oscillation decreases exponentially with the retarded time (i.e. the time at which the wave started from the source.). Thus we write for the wave amplitude.

This means that

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.185. A plane longitudinal harmonic wave propagates in a medium with density ρ. The velocity of the wave propagation is v. Assuming that the density variations of the medium, induced by the propagating wave, Aρ  ≪  ρ, demonstrate that
 (a) the pressure increment in the medium 
Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev where Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev is the relative deformation;
 (b) the wave intensity is defined by Eq. (4.3i). 

Ans. (a) Let us consider the motion of an element of the medium of thickness dx and unit area of cross-section. Let Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev displacement of the particles of the medium at location x. Then by the equation of motion

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

where dp is the pressure increment over the length dx

Recalling the wave equation

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

we can write the foregoing equation as

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Integrating this equation, we get

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

In the absence of a deformation (a wave), the surplus pressure is Δp = 0. So ’Const’ = 0 and

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(b) We have found earlier that

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

It is easy m see that the space-time average of both densities is the same and the space time average of total energy density is then

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

The intensity of the wave is

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Using

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.186. A ball of radius R = 50 cm is located in the way of propagation of a plane sound wave. The sonic wavelength is λ = 20 cm, the frequency is v = 1700 Hz, the pressure oscillation amplitude in air is (Δp)= 3.5 Pa. Find the mean energy flow, averaged over an oscillation period, reaching the surface of the ball. 

Ans. The intensity of the sound wave is  

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Thus the mean eneigy flow reaching the ball is

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRevbeing the effective area (area of cross section) of the ball. Substitution gives 10.9 mW.

 

Q.187. A point A is located at a distance r = 1.5 m from a point isotropic source of sound of frequency v = 600 Hz. The sonic power of the source is P = 0.80 W. Neglecting the damping of the waves and assuming the velocity of sound in air to be equal to v = 340 m/s, find at the point A:
 (a) the pressure oscillation amplitude (Δp)m and its ratio to the air pressure;
 (b) the oscillation amplitude of particles of the medium; compare it with the wavelength of sound. 

Ans.

 Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

(b) We have

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

 

Q.188. At a distance r = 100 m from a point isotropic source of sound of frequency 200 Hz the loudness level is equal to L = 50 dB. The audibility threshold at this frequency corresponds to the sound intensity l0  = 0.10 nW/m2. The damping coefficient of the sound wave is γ = 5.0.10-4  m-1. Find the sonic power of the source.

Ans.  Express L in bels. (i.e. L = 5 bels).

Then the intensity at the relevant point (at a distance r from the source) is Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Had there been no damping the intensity would have been Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Now this must equal the quantity 

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRevwhere P = sonic power of the source.

Thus Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

or  Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Sample Paper

,

Extra Questions

,

Objective type Questions

,

MCQs

,

Free

,

pdf

,

past year papers

,

mock tests for examination

,

study material

,

video lectures

,

Semester Notes

,

practice quizzes

,

Exam

,

ppt

,

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

Summary

,

Important questions

,

Viva Questions

,

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

,

Irodov Solutions: Elastic Waves. Acoustics- 2 Notes | EduRev

;