How Do Doppler Effects Determine the Frequency of a Whistle in Motion?

[Jacob87411]

A Boy is walking away from a wall at a speed of 1m/s in a direction at right angles to the wall. As he walks, he blows a whistle steadily. An observer toward whom the boy is walking hearts 4 beats per second. If the speed of sound is 340 m/s, what is the frequency of the whistle?

My original approach did not come out anywhere near right, so I am just going to ask if this analysis of the situation is right.

The boy is walking toward someone blowing a whistle. So there are two doppler effects? The one caused from the boy walking toward an observer and the sound wave reflecting off the wall, so I would calculate two doppler effects and combine them? And how can I change beat frequency to frequency, if that's possible even.

 

[Gokul43201]

The observer will hear two different frequencies, as you guessed correctly : one coming directly from the whistler (doppler shifted upwards) and another from the reflection off the wall (doppler shifted downwards). If the whistle frequency is f, you can find f1 (>f) and f2(<f) in terms of f, using the doppler relation. The beat frequency is simply the difference between these two frequencies. So, f1 - f2 = 4 Hz.

You now have 3 equations in 3 unknowns, which you can solve to find them all.

 

[vincentchan]

Hints:
The guy hears two whistle, one is moving toward the guy at 1m/s and the other one is moving away, now you know the frequency of two whistle, and ...(I don't want to do the problem for you)

 

[Jacob87411]

Question..this is the doppler effect equation for the source moving towards a stationary observer:

F'= F / (1-Vsource/V)

VSource = 1 m/s
V= 340 m/s for sound of speed

What goes in for F? 4? THats what I did originally but it didnt work

 

[Jacob87411]

Or will the two doppler effects be equal to each other and that's how you find the F?