Question

A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz.
(a) What overall change in requency is detected by a person on the platform as the train moves from approaching to receding?
(b) What wavelength is detected by a person on the platform as the train approaches?

Answer 1

Answer:

a) -75 Hz

b) 0.11 $m^{-1}$

Explanation:

a) Let us first find the frequency detected by the person on the platform.

We have to find the frequency observed by the person when the train was approaching and when the train was receding.

When the train was approaching:

$f_o = \frac{v}{v - v_s} f_s$

where fo = frequency observed

fs = frequency from the source = 320 Hz

v = speed of sound = 343 m/s

vs = speed of the train = 40 m/s

Therefore:

$f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz$

The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.

When the train was receding:

$f_o = \frac{v}{v + v_s} f_s$

$f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz$

The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.

Therefore, the frequency change is given as:

Δf = 287 - 362 = -75 Hz

b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:

$v = \lambda f$

where λ = wavelength

f = frequency of the train as it approaches = 362 Hz

v = speed of train = 40 m/s

Therefore, the wavelength detected is:

40 = λ * 362

λ = 40 / 362 = 0.11 $m^{-1}$