Question

A train running at 108 km h−1 towards east whistles at a dominant frequency of 500 Hz. Speed of sound in air is 340 m/s. What frequency will a passenger sitting near the open window hear? (b) What frequency will a person standing near the track hear whom the train has just passed? (c) A wind starts blowing towards east at a speed of 36 km h−1. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.

Answer 1

The frequency heard by passenger f = 500 Hz , near the track f = 460 Hz and speed of wind = 10 m/s.

Explanation:

We are given that:

• Speed of train Vs = 108 km / h = 108 x 5 /18 = 30 m/sec
• Frequency = 500 Hz
• Speed of sound = 340 m/s

(a) Frequency near the track:

f = fo [V sound ± Vob / Vsound ±Vs]

f = 500 [ 340 + 30 / 340 + 30 ]

f = 500 Hz

(b) Vob = 0

f = fo [V sound / Vsound ±Vs]

f = 500 x 340 / 340 + 30

f = 459 Hz

(c) Speed of wind V(w) = 36 km/h

V(w) = 36 x 5 / 18 = 10 m/s

Frequency heard by the person:

n = v + v0 / v + vs × n0

n = 340 + 0 / 340 + 30 × 500 = 459 Hz

Frequency near the track:

n = ( V(sound) + V(w) / (v + v(air) + vs) × n0  

=(340 + 10)(340 + 10 +30) × 500

=460 Hz

Hence the frequency heard by passenger f = 500 Hz , near the track f = 460 Hz and speed of wind = 10 m/s.