C Exercise 14.51

Part A = f_{o} = 426 Hz

Part B = f_{o} = 377 Hz

Solutions Below:

Part A

First, find the velocity of sound in air.

The speed of sound is given by (approximately):

v_{sound} = 331.4 * sqrt((T_{C} + 273.15) / 273.15)

Since the temperature is 25 °C, the speed of sound is about 346.234m/s

Now we can use the formula for the Doppler effect to find out the frequency, f_{o} that the observer hears:

f_{o} = f_{actual}*[(v_{sound} + v_{observer}) / (v_{sound} + v_{source}))

The formula is from the perspective of the observer moving left to right and the source moving from right to left (if both are approaching each other).

First, convert the speed of the train to m/s from km/h:

75,000km/h = 20.833m/s

Now solve:

f_{o} = f_{actual}*[(v_{sound} + v_{observer}) / (v_{sound} + v_{source}))

f_{o} = 400*[(346.234 + 0) / (346.234 - 20.833))

f_{o} = 400*[346.234 / 325.401]

f_{o} = 426 Hz

Part B

For this problem, solve using the same method as Part A, except now the train is moving away from the observer instead of towards it (so the sign changes and the velocity is +20.833m/s instead of -20.833m/s):

f_{o} = f_{actual}*[(v_{sound} + v_{observer}) / (v_{sound} + v_{source}))

f_{o} = 400*[(346.234 + 0) / (346.234 + 20.833))

f_{o} = 400*[346.234 / 367.067]

f_{o} = 377 Hz