# Calculating moving speaker through two detectors

• Samuelriesterer
The frequency of the light blinking on each of the detectors is##f = \frac{v}{\lambda} = 4.857 X 10^8 Hz##f

## 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?

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.

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?

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.

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##