Chapter 15 Waves Solutions
Question - 21 : - A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s–1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s–1? The speed of sound in still air can be taken as 340 m s–1.
Answer - 21 : -
For the stationary observer: 400 Hz; 0.875 m; 350 m/s
For the running observer: Not exactly identical
For the stationary observer:
Frequency of the sound produced by the whistle, ν = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, v =10 m/s
As there is no relative motion between the source and theobserver, the frequency of the sound heard by the observer will be the same asthat produced by the source, i.e., 400 Hz.
The wind is blowing toward the observer. Hence, theeffective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, ve =340 + 10 = 350 m/s
The wavelength (λ) of the sound heardby the observer is given by the relation:
For the running observer:
Velocity of the observer, vo =10 m/s
The observer is moving toward the source. As a result ofthe relative motions of the source and the observer, there is a change infrequency (
).
This is given by the relation:
Since the air is still, the effective speed of sound = 340+ 0 = 340 m/s
The source is at rest. Hence, the wavelengthof the sound will not change, i.e., λ remains 0.875 m.
Hence, the given two situations are not exactly identical.
Question - 22 : - A travelling harmonic wave on a string is described by
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
Answer - 22 : -
(a) The given harmonic wave is:
For x = 1 cm and t = 1s,
The velocity of the oscillation at a given point and timeis given as:
Now, the equation of a propagating wave is given by:
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:
Therefore, all the points at distances nλ
, i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.
Question - 23 : - A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 
or 0.05 Hz?
Answer - 23 : -
(a) (i)No
(ii)No
(iii)Yes
(b) No
Explanation:
(a) The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.
(b) The short pip produced after every 20 s does not mean that the frequency of the whistle is
or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.
Question - 24 : - One end of a long string of linear massdensity 8.0 × 10–3 kg m–1 is connected toan electrically driven tuning fork of frequency 256 Hz. The other end passesover a pulley and is tied to a pan containing a mass of 90 kg. The pulley endabsorbs all the incoming energy so that reflected waves at this end havenegligible amplitude. At t = 0, the left end (fork end) of thestring x = 0 has zero transverse displacement (y =0) and is moving along positive y-direction. The amplitude of thewave is 5.0 cm. Write down the transverse displacement y asfunction of x and t that describes the waveon the string.
Answer - 24 : -
The equation of a travelling wavepropagating along the positive y-direction is given by thedisplacement equation:
y (x, t)= a sin (wt – kx) … (i)
Linear mass density, 
Frequency of the tuning fork, ν = 256 Hz
Amplitude of the wave, a =5.0 cm = 0.05 m … (ii)
Mass of the pan, m = 90 kg
Tension in the string, T = mg= 90 × 9.8 = 882 N
The velocity of the transverse wave v,is given by the relation:
Substituting the values from equations (ii),(iii), and (iv) in equation (i), we get the displacementequation:
y (x, t)= 0.05 sin (1.6 × 103t – 4.84 x)m
Question - 25 : - A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h–1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s–1.
Answer - 25 : -
Operating frequency of the SONAR system, ν = 40 kHz
Speed of the enemy submarine, ve =360 km/h = 100 m/s
Speed of sound in water, v =1450 m/s
The source is at rest and the observer (enemy submarine)is moving toward it. Hence, the apparent frequency () received and reflected by thesubmarine is given by the relation:
The frequency (
) received by the enemy submarineis given by the relation:
Question - 26 : - Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s–1, and that of P wave is 8.0 km s–1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer - 26 : -
Let vSand vP bethe velocities of S and P waves respectively.
Let L be the distancebetween the epicentre and the seismograph.
We have:
L = vStS (i)
L = vPtP (ii)
Where,
tS and tP arethe respective times taken by the S and P wavesto reach the seismograph from the epicentre
It is given that:
vP = 8 km/s
vS = 4 km/s
From equations (i) and (ii),we have:
vS tS = vP tP
4tS = 8 tP
tS = 2 tP (iii)
It is also given that:
tS – tP =4 min = 240 s
2tP – tP =240
tP = 240
And tS =2 × 240 = 480 s
From equation (ii), we get:
L = 8 ×240
= 1920 km
Hence, the earthquake occurs at a distance of 1920 km fromthe seismograph.
Question - 27 : - A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer - 27 : -
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
Velocity of the bat, vb = 0.03 v
Where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as:
This frequency is reflected by the stationary wall (
) toward the bat.The frequency (
) of the received sound is given by the relation: