Calculating moving speaker through two detectors

[Samuelriesterer]

Homework Statement

A speaker emitting sound at a frequency of 20 Hz is moving in the +x direction between two detectors. The speaker is moving at a speed of 30 m/s and the detectors are wired so that they flash red (λ = 700 nm) when the pressure is a maximum and green (λ = 700 nm) when the pressure is a minimum. Use 340 m/s as the speed of sound.

[1] Calculate the distance between maxima ahead and behind the moving speaker.

[2] Calculate the frequency of the light blinking on each of the detectors.

[3] An observer between the detectors is moving at 0.8c in the +x direction. Calculate the frequencies for the red and green lights for each detector this observer sees.

[4] Transform the velocity of the speaker to the observer frame as well as the velocity of sound in each direction.

2. Homework Equations
$f' = f \frac{v \pm v_{receiver}}{v \pm v_{source}}$

$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$

The Attempt at a Solution

I am confused at the start with how the detectors detect maximum and minimum pressure and subsequently question1 about the maxima?

 

[Simon Bridge]

How the detectors work does not matter.
Sound waves are pressure waves - the distance between two consecutive maxima of a wave has a special name.

 

[Samuelriesterer]

OK thanks, wavelength!

[1] Calculate the distance between maxima ahead and behind the moving speaker.

$\lambda_{behind} = \frac{v_{sound} + v_{source}}{f} = 18.5 m$
$\lambda_{ahead} = \frac{v_{sound} - v_{source}}{f} = 15.5 m$

[2] Calculate the frequency of the light blinking on each of the detectors.

$f = \frac{v}{\lambda} = 4.857 X 10^8 Hz$

[3] An observer between the detectors is moving at 0.8c in the +x direction. Calculate the frequencies for the red and green lights for each detector this observer sees.

$T = \frac{v_{speaker} - v_{observer}}{1 - \frac{v_{speaker} v_{observer}}{c^2}} = 2.0588 X 10^{-9} s$

$T' = \frac{T}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{2.0588 X 10^{-9} s}{\sqrt{1 - \frac{(.8c)^2}{c^2}}} = 3.4314 X 10^{-9} s$

$f' = \frac{1}{T'} = 2.9143 X 10^8 s$

[4] Transform the velocity of the speaker to the observer frame as well as the velocity of sound in each direction.

$v'_{speaker} = \frac{v_{speaker} - v_{observer}}{1 - \frac{v_{speaker} v_{observer}}{c^2}} = -2.3983486 X 10^8 m/s$

$v'_{sound +x direction} = \frac{v_{sound +x direction} - v_{observer}}{1 - \frac{v_{sound +x direction} v_{observer}}{c^2}} = \frac{340 m/s - .8c}{1 - \frac{(340 m/s)(.8c)}{c^2}} = -2.3985502 X 10^8 m/s$

$v'_{sound -x direction} = \frac{v_{sound +x direction} - v_{observer}}{1 - \frac{v_{sound +x direction} v_{observer}}{c^2}} = \frac{-340 m/s - .8c}{1 - \frac{(-340 m/s)(.8c)}{c^2}} = -2.39812182 X 10^8 m/s$

What is throwing me off now is that the velocity of sound in the -x direction according to the observer's frame is moving slower than the sound moving in the +x direction. I would think that it should be the opposite?

 

[Simon Bridge]

In the observers reference frame, the detectors are moving in the -x direction at 0.8c ... one of the detectors is behind.
You don't need the velocity of sound in the observer's frame to do the problem - only the frequency of the flashing lights in their rest frame and the relative velocities.

 

[Samuelriesterer]

Oh I see:

$v'_{ahead} = v_{ahead} - v_{observer} = \frac{f}{\lambda} - .8c = \frac{7X10^{-7}}{15.5} - .8c = -2.398335999999999548 X10^8 m/s$

$v'_{behind} = -v_{behind} - v_{observer} =- \frac{f}{\lambda} - .8c = -\frac{7X10^{-7}}{18.5} - .8c = -2.398336000000000378 X10^8 m/s$