Question:

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:

1. Calculate the speed of the P wave: 8.0 km s^−1

2. Calculate the time difference between the arrival of the P and S waves: 4 min

3. Convert the time difference to seconds: 4 min = 240 s

4. Calculate the distance traveled by the P wave in the given time: 8.0 km s^−1 x 240 s = 1920 km

5. The distance traveled by the P wave is equal to the distance of the earthquake from the seismograph: 1920 km

Question:

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 1/20 or 0.05 Hz?

Answer:

a) (i) Yes, the pulse has a definite frequency. (ii) Yes, the pulse has a definite wavelength. (iii) Yes, the pulse has a definite speed of propagation.

b) No, the frequency of the note produced by the whistle is not equal to 1/20 or 0.05 Hz. The frequency of the note produced by the whistle is equal to 1 Hz.

Question:

One end of a long string of linear mass density 8.0 ×10^−3 kg m^−1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string :

Answer:

y(x,t) = 5 cm * sin (2π256 Hzt - 2π*x/λ)

Question:

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:

1. The speed of sound in air is calculated as follows: Speed of sound in air = 343 m/s

2. The speed at which the bat is moving is calculated as follows: Speed of the bat = 0.03 x 343 m/s = 10.29 m/s

3. The frequency of the sound reflected off the wall is calculated as follows: Frequency of the sound reflected off the wall = (Speed of sound in air - Speed of the bat) / 2 = (343 m/s - 10.29 m/s) / 2 = 166.355 kHz

Therefore, the frequency that the bat hears reflected off the wall is 166.355 kHz.

Question:

A travelling harmonic wave on a string is described by y(x,t)=7.5sin(0.0050x+12t+π/4)
(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=2s , 5 s and 11s

Answer:

(a) Displacement: y(1,1) = 7.5sin(0.0050(1)+12(1)+π/4) = 5.4 cm

Velocity: y’(1,1) = 7.5(0.0050)cos(0.0050(1)+12(1)+π/4) = 0.37 cm/s

No, this velocity is not equal to the velocity of wave propagation. The velocity of wave propagation is the speed of the wave, which is the rate at which the wave moves through the medium.

(b) Points with same displacement and velocity as x=1 cm point at t=2s:

y(x,2) = 7.5sin(0.0050x+12(2)+π/4)

y’(x,2) = 7.5(0.0050)cos(0.0050x+12(2)+π/4)

Points with same displacement and velocity as x=1 cm point at t=5s:

y(x,5) = 7.5sin(0.0050x+12(5)+π/4)

y’(x,5) = 7.5(0.0050)cos(0.0050x+12(5)+π/4)

Points with same displacement and velocity as x=1 cm point at t=11s:

y(x,11) = 7.5sin(0.0050x+12(11)+π/4)

y’(x,11) = 7.5(0.0050)cos(0.0050x+12(11)+π/4)

Question:

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:

1. Convert the speed of the enemy submarine from km/h to m/s: 360 km/h = 100 m/s

2. Calculate the Doppler shift of the frequency: Doppler shift = (speed of sound + speed of enemy submarine) / (speed of sound - speed of enemy submarine) Doppler shift = (1450 m/s + 100 m/s) / (1450 m/s - 100 m/s) Doppler shift = 43.5

3. Calculate the frequency of the sound reflected by the submarine: Frequency of sound reflected = Doppler shift x original frequency Frequency of sound reflected = 43.5 x 40.0 kHz Frequency of sound reflected = 1740 kHz