Question

You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 98 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 76 Hz. Take the speed of sound in air to be v = 340 m/s(a) Find an equation for the speed of the sound source v., in this case it is the speed of the train. Express your answer in terms of f2. and v.

Answer 1

Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

Collecting like terms

vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

v = (33320 - 340f2)/( f2 + 98)