Doppler shift above speed of sound.

[Cypripedium]

Homework Statement

A source and an observer are each traveling at 0.52 times the speed of sound. The source emits sound waves at 1.2 kHz.

(a) Find the observed frequency (in kHz) if the source and observer are moving toward each other.

(b) Find the observed frequency (in Hz) if the source and observer are moving away from each other.

Homework Equations

frequency observed (f_o) = (1 / (1 - (V_source/V_sound))) * f_source

Change V_source to negative if moving away from each other.

The Attempt at a Solution

I'm assuming we're going with 340 m/s for the speed of sound, but it shouldn't even matter because it'll always be a 1.04 ratio in the equation.

For part (a): Moving toward each other, so we can act like the observer is stationary and the source is moving at him at 1.04 times the speed of sound, 353.6 m/s. Plugging that into the equation and get -30 kHz. I don't know if that's actually incorrect, the negative just scares me.

For part (b): Moving away from each other, so we can act like the observer is stationary and the source is moving away from him at 1.04 times the speed of sound, 353.6 m/s. Plugging that into the equation gets me 588.24 Hz. I don't actually know if that is incorrect though.

Am I doing these correctly? I feel like the standard equation shouldn't really apply above the speed of sound, but what do I know, I'm just in an intro physics class! Thanks for the help.

 

[dwintz02]

You should not get a negative frequency. Without calculating, I believe you are close. The actual values of your answers make sense, the frequency should become much higher when they are moving towards each other (picture waves being 'scrunched' together) and the answer should be lower when they are moving away relative to each other (picture waves being 'stretched' out).

 

[Shooting Star]

When both source and sound are moving, you should use this formula:

f’ = f(v+-vo)/(v+-vs), where v is the speed of sound. Choose + or - signs to suit the particular case.