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.

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)

answer is slate

hope this