Speed of sound in liquid given Doppler shift in air and liquid

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tmkgemini
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A microphone is moving in air toward a stationary source of sound (speed of sound = 343m/s). The detected frequency is 82.7Hz greater than the emitted frequency. When the microphone moves at the same speed toward the same stationary source in a liquid, the detected frequency is only 21.6Hz greater than the emitted frequency. What is the speed of sound in the liquid?


I really need help! I can't figure out this problem and I've gone all through my textbook and lecture notes!
 
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tmkgemini said:
A microphone is moving in air toward a stationary source of sound (speed of sound = 343m/s). The detected frequency is 82.7Hz greater than the emitted frequency. When the microphone moves at the same speed toward the same stationary source in a liquid, the detected frequency is only 21.6Hz greater than the emitted frequency. What is the speed of sound in the liquid?

This is just doppler shift. Find the speed of the microphone using the doppler formula. Then set up the equation for doppler shift in the second situation. You know the doppler shift and the relative speeds so the only unknown is the speed of sound in the liquid.

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Calculating the speed of sound in a liquid can be a tricky problem, but with the right approach, it can be solved easily. Let's break down the given information and use the speed of sound formula to find the speed of sound in the liquid.

First, we know that the speed of sound in air is 343m/s. This will be our initial velocity, or v1. The detected frequency in air is 82.7Hz greater than the emitted frequency, so we can use the formula f = v/λ to find the wavelength (λ) of the sound wave in air. Since the microphone is moving towards the stationary source, we can use the relative velocity formula v2 = v1 + v, where v is the velocity of the microphone. This will give us the velocity of the sound wave in air as it reaches the microphone.

Now, let's look at the situation in the liquid. The only difference is that the detected frequency is only 21.6Hz greater than the emitted frequency. Using the same formula, f = v/λ, we can find the wavelength (λ) of the sound wave in the liquid. Again, using the relative velocity formula, we can find the velocity of the sound wave in the liquid as it reaches the microphone.

Now, we have two equations with two unknowns (v and λ). We can solve for the speed of sound in the liquid by setting these two equations equal to each other and solving for v. This will give us the velocity of the microphone, which is also the velocity of the sound wave in the liquid.

So, in summary:

v1 = 343m/s (speed of sound in air)
f1 = emitted frequency
f2 = detected frequency
λ1 = wavelength in air
λ2 = wavelength in liquid
v2 = velocity of sound in liquid

Using the formula f = v/λ, we can set up the following equations:

f2 = (v1 + v) / λ1
f2 = (v2 + v) / λ2

Equating these two equations and solving for v, we get:

(v1 + v) / λ1 = (v2 + v) / λ2
v1λ2 + vλ2 = v2λ1 + vλ1
v1λ2 - vλ1 = v2λ1 - vλ2
v = (v2λ1 - v1λ2