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 = 92 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 79 Hz. Take the speed of sound in air to be v = 340 m/s.

Randomized Variables
fi = 92 Hz
f2 = 77 Hz

Find an equation for the speed of the sound source vs, in this case it is the speed of the train.

Answer 1

Equations for the speed of the train

340 – vs = 3.70fs

340 + vs = 4.42fs

Ans solving simultaneously,

Vs = 29.2m/s

Step-by-step explanation:

Please see the attachments below.

Answer 2

Given Information:

Frequency of whistle as train approaches = f₁ = 92 Hz

Frequency of whistle as train recedes = f₂ = 79 Hz

Speed of sound = v = 340 m/s

Required Information:

equation for speed of train = ?

Speed of train = v₀ = ?

Answer:

92 = f₀ ( 340/(340 + v₀) )

79 = f₀ ( 340/(340 - v₀) )

Speed of train = v₀ = 25.76 m/s

Explanation:

There are two cases here with respect to the movement of train and standing person

Case 1: When the train approaches

f₁ = f₀ ( v/( v + v₀) )

Where f₀ is the frequency of the whistle when train crosses the person

92 = f₀ ( 340/(340 + v₀) )

Case 2. When the train recedes

f₂ = f₀ ( v/( v - v₀) )

79 = f₀ ( 340/(340 - v₀) )

We have two equations and two unknowns, we can simultaneously solve these two equations,

92 = f₀ ( 340/(340 + v₀) ) eq. 1

79 = f₀ ( 340/(340 - v₀) ) eq. 2

Dividing eq. 1 by eq. 2, f₀ cancels out

92/79 = ( 340/(340 + v₀) )/( 340/(340 - v₀) )

340 m/s cancels out

1.164 = (340 + v₀) / (340 - v₀)

1.164(340 - v₀) = (340 + v₀)

395.76 - 1.164v₀ = 340 + v₀

395.76 - 1.164v₀ - 340 - v₀ = 0

-2.164v₀ + 55.76 = 0

-2.164v₀ = -55.76

v₀ = 55.76/2.164

v₀ = 25.76 m/s

Therefore, the speed of the train (sound source) is 25.76 m/s.