Problem 1

1.) (10 points) A tuning fork with a frequency 440 Hz is dropped from a tower of height 100m by a student. She can hear the fork as it falls. Given that the speed of sound in air is v_sound = 330 m/s, what frequency does the student hear as a function of time? Do not forget that it takes time for the sound to travel back to the student. Take t = 0 to be when the student lets go of the fork.

 

$f' = f (\frac{v \pm v_{O}}{v \mp v_{S}})$

 

Answer:
The doppler shift equation reduces to
$f' = f(\frac{v}{v + v_{s}})$
since the observer is stationary and the source is moving away from the observer. Now, we have to find the velocity of the source. If we neglect the travel time of the sound back to the source, we have
$f' = f(\frac{v}{v + gt})$
However, we have to take into account the travel time for the sound back to the observer. If we consider the time that it takes for the fork to fall a distance d to be t₁, and the time it takes for the sound to reach the observer after leaving the tuning fork to be t₂, then the total time is given by
t = t₁ + t₂
After a time t has gone by for the observer, she hears the frequency that the fork was emitting at a time t₁.
$f'(t) = f(\frac{v}{v + g t_{1}})$
But, t₁ = t - t₂.
$f'(t) = f(\frac{v}{v + g(t - t_{2})})$
So, what is t₂? That is the time that it takes for the sound to travel a distance d.
$t_{2} = \frac{d}{v}$
So, we have
$f'(t) = f(\frac{v}{v + g(t - \frac{d}{v})})$
Now, since d depends on t, we need to solve for it in terms of t. The fork falls a distance d given by
$d = v_{0}t + \frac{1}{2} g t^{2}$
$d = \frac{1}{2} gt^{2}$
So, that gives us
$f'(t) = \frac{v}{v + g(t - \frac{g t^{2}}{2})}$
Putting in the numbers, we find
$f'(t) = (440)(\frac{330}{300 + 9.8 t - .146 t^{2}})$