Question

A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle is 420 Hz.
1. What is the wavelength of the sound waves in front of the locomotive? Use 344 m/s for the speed of sound in air.
2. What is the wavelength of the sound waves behind the locomotive? Use 344 m/s for the speed of sound in air.
3. What is the frequency of the sound heard by a stationary listener in front of the locomotive? Use 344 m/s for the speed of sound in air.
4. What is the frequency of the sound heard by a stationary listener behind the locomotive? Use 344 m/s for the speed of sound in air.

Answer 1

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz