Calculations for the Doppler Effect with a Movin Observer

We will now consider the above case in the form of a specific problem. We are once again given an original frequency emitted by some source of sound and told to calculate the observed frequency. Observer approaching the source Suppose the bank alarm is emitting a sound at a frequency of 500Hz and we are driving toward it at 22 m/s. What is the frequency that we hear? Once again, we are looking for the observed frequency, f . We will again use our equation:
f' = v'/L'
All we have to do is find v'and L' and we can get our answer. Both are pretty easy to find. Since the car is moving into the sound waves, the velocity of the waves relative to the car will be greater. We simply add the velocity of the car to the speed of sound:
v' = 343 + 22 = 365 m/s
Because the bank is not moving, the sound waves will not be bunched up or spread out. The observed wavelength, L', will be the same as the original wavelength L . We can find this from the given information in the problem:
L' = L = v/f = 343/500 = 0.686m
We use these two numbers, v' and L' , to find the observed frequency:
f' = v'/L' = 365/0.686 = 532 Hz
So we see that when the observer is approaching the source, he or she hears a higher frequency than when still. If that example made sense to you, it should now be easy understand the next case.