A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?

First, we need to find the frequency of the sound wave when it reaches the reflector. We will use the Doppler effect formula for moving source and moving observer: $$f' = f \frac{v \pm v_o}{v \pm v_s}$$ where \(f'\) is the observed frequency, \(f\) is the source frequency, \(v\) is the speed of sound in the medium, \(v_o\) is the speed of the observer (reflector) relative to the medium, \(v_s\) is the speed of the source relative to the medium, and the signs depend on the motion direction. In this case, since the source is moving toward the reflector and the reflector is moving toward the source, we can use the positive signs: $$f' = f \frac{v + v_o}{v - v_s}$$ In our example, we have \(f = 100.0 \thinspace \text{Hz}\), \(v = 343 \thinspace \text{m/s}\) (speed of sound in air at room temperature), \(v_o = 5.00 \thinspace \text{m/s}\), and \(v_s = 10.00 \thinspace \text{m/s}\). Now we can find \(f'\): $$f' = 100 \thinspace \text{Hz} \frac{343+5}{343-10} \Rightarrow f' \approx 109.1 \thinspace \text{Hz}$$ So, the frequency of the sound wave at the reflector is approximately \(109.1 \thinspace \text{Hz}\).

Now that we have the frequency of the sound wave at the reflector, we can use the Doppler effect formula again to find the frequency of the reflected sound wave as it is detected by the listener back at the source. This time, since the listener is static, and the reflector is moving to the left, we will use a '+' for \(v_o\) and a '-' for \(v_s\) in the formula: $$f'' = f' \frac{v + v_s}{v - v_o}$$ With \(f' = 109.1 \thinspace \text{Hz}\), \(v = 343 \thinspace \text{m/s}\), \(v_o = -5.00 \thinspace \text{m/s}\), and \(v_s = -10.00 \thinspace \text{m/s}\), we can find the frequency \(f''\): $$f'' = 109.1 \thinspace \text{Hz} \frac{343 - 5}{343 + 10} \Rightarrow f'' \approx 99.5 \thinspace \text{Hz}$$ So, the frequency of the reflected sound wave detected by the listener back at the source is approximately \(99.5 \thinspace \text{Hz}\).