How Does Doppler Effect Influence the Frequency of a Moving Siren and Its Echo?

[grouper]

Homework Statement

A siren emitting a sound of frequency 1000 Hz moves away from you towards the face of a cliff at a speed of 10 m/s. Take the speed of sound in air to be 330 m/s.

a. What is the frequency of the sound you hear directly from the siren?

b. What is the frequency of the sound you hear reflected off the cliff?

c. What is the beat frequency between the two sounds?

Homework Equations

f(BEATS)=f1-f2

The Doppler Effect
f'=f((v(air)+-v(detector))/(v(air)+-v(source)))

The Attempt at a Solution

The beat frequency is easy, but I am confused on how to use the Doppler Effect equation (i.e. which signs to use, etc). Thanks for the help!

 

[Oddbio]

You can find the correct sign convention by realizing that if the (net) velocity of the source and listener is toward each other, then the frequency should increase.

So the sign convention is to first look at the direction from the listener toward the source. Take this direction to be positive. That's it. and remember to take their velocities relative to the medium.

 

[grouper]

Well, since the detector is not moving, the velocity of the detector is 0 m/s and since you need to calculate the frequency of the siren both as it moves away from you and as it moves towards the cliff, I went ahead and used both +10 m/s and then -10 m/s in the denominator. This gave me 970.6 Hz, which I assume would be answer (a) because the frequency should decrease if it is moving away, correct? And 1031 Hz would be the frequency of the sound bouncing off the cliff because it should increase as it moves towards the cliff, correct? These two frequencies result in a beat frequency of 60.4 Hz. I believe I figured it out, but if someone could please make sure I used the Doppler Effect equation correctly and reassure me that I did this correctly it would be greatly appreciated. Thank you.

 

[Oddbio]

It all looks good to me :)

But, while there is nothing wrong with the way you did it, your logic is absolutely correct, it might benefit you to learn the sign convention for general situations in which such reasoning might not work.

The direction from the listener to the source is the positive direction. Therefore the source is moving at +10m/s, and that is the observed frequency directly from the source.
However, when the waves bounce off the cliff is is equivalent to having the source change direction, and now it is moving in a direction opposite from listener->source (because it's going from source->listener). So in the second scenario the velocity is taken as -10m/s.