Question

At a factory, a noon whistle is sounding with a frequency of 480 Hz. As a car traveling at 85 km/h approaches the factory, the driver hears the whistle at frequency fi. After driving past the factory, the driver hears frequency ff. What is the change in frequency ff − fi heard by the driver? (Assume a temperature of 20° C.)

I tried this , but its wrong
[2(343)(22.22) / ((343)² - (22.22)²)] (480)= 62.27 Hz

COuld someone help me out please.

Answer 1

The received frequency is higher (compared to the emitted frequency) during the approach, it is identical at the instant of passing by, and it is lower during the recession.
f1=f(o)•(1+v/V),
f2=f(o)•(1-v/V),
v = 85 km/h = 23.61 m/s,
V =343 m/s,
Δf =f1-f2 = f(o)•(1+v/V)-f(o)•(1-v/V)= =2•f(o)•(v/V) =
= 2•480•23.61/343 =66.08 Hz

Answer 2

I cant solve this can someone help me please