Calculating Well Depth with Doppler & Echo Effects

[tigermonica]

Homework Statement

An alarm clock has been thrown down a well.

around 100m deep well
an alarm clock with a frequency of 368hz sounds for 3 seconds
accelerating at 9.8 m/s2
initial velocity 0

Homework Equations

How deep is the well? using doppler
take in consideration of the echo
what is the frequency of beats?

The Attempt at a Solution

I really have no idea :s
all i got was that it had red shift and blue shift.
the wall is the observer?

 

[AvroArrow]

the frequency of the beats would be the sum of the initial frequency and the frequency of the echo? Would the frequency decrease due to the redshift?

 

[kerimek]

I can provide a possible solution to this problem by using the principles of Doppler and echo effects. First, we need to understand that the frequency of sound waves changes when the source of the sound is moving towards or away from the observer. In this case, the alarm clock is moving towards the bottom of the well, which will cause a decrease in frequency (red shift) due to the Doppler effect.

We can use the formula for Doppler effect, f'=f(v±vo)/(v±vs), where f' is the observed frequency, f is the original frequency, v is the speed of sound, vo is the velocity of the observer (in this case, the wall), and vs is the velocity of the source (in this case, the alarm clock). Since the initial velocity is 0, we can ignore that term and use the formula f'=f(v-vo)/v.

Next, we need to take into account the echo effect. This means that the sound waves will bounce off the bottom of the well and travel back up, causing an increase in frequency (blue shift). We can use the formula f"=f'(2L/v), where f" is the final frequency after the echo, f' is the frequency after the Doppler effect, L is the depth of the well, and v is the speed of sound.

Now, we can combine these equations to solve for the depth of the well, L. We know that the original frequency is 368 Hz, the final frequency after the echo is 368 Hz, and the speed of sound is approximately 343 m/s. Plugging in these values, we get 368=f(343-vo)/343 and 368=f(2L/343). Solving for vo and L, we get vo=343-368= -25 m/s and L=343/2=171.5 m.

Therefore, the well is approximately 171.5 meters deep. This solution takes into account both the Doppler and echo effects, which can affect the frequency and give us a more accurate measurement of the well's depth. Additionally, we can also calculate the frequency of beats by using the formula fbeat=|f'-f"|, which in this case would be 0 Hz. This means that there are no beats, as the final frequency after the echo is equal to the original frequency.