Doppler Effect, conceptual problems (3)

[Mayy]

Homework Statement

a. A fire engine is moving at 40 m/s and sounding its horns. A car in front of the fire engine is moving at 30 m/s, and a van in front of the car is stationary. Which observer hears the fire engine's horn at higher pitch, the driver of the car or the van?

b. A bat flying toward a wall emits a chirp at 40 kHz. Is the frequency of the echo received by the bat greater than, less than, or equal to 40 kHz? Explain.

c. Two tuning forks are struck. One at 748 Hz and the other at 234 Hz. Which wavelength travels faster? (this one is actually reworded differently, I didn't write down the problem, but it was something along these lines...)

Homework Equations

wavelength = 4L

I don't think any are required for these?

The Attempt at a Solution

Well I know that the closer the wavelengths are, the higher the pitch is. However, for (a), there are two moving objects, which throw me off.

And for (b), since the bat is still moving towards the wall, does that have any effect?

As for (c), would they both travel at the same speed (speed of sound)?

 

[shallgren]

The equation you're missing is:

[tex]f_o = f_s \cdot \frac{v \pm v_o}{v \mp v_s} [/tex]

Where [tex]f_o[/tex] is the observed frequency
[tex]f_s[/tex] is the original frequency
[tex]v[/tex] is the speed of sound
[tex]v_o[/tex] is the speed of the observer
[tex]v_s[/tex] is the speed of the object producing the sound.

The hardest part is getting the signs right with two moving objects. The best way of doing this is consider what would happen if each one of the objects was stationary in separate scenarios. The example my text has for this is a submarine traveling at 5 m/s detecting a signal from a submarine behind it traveling at an unknown speed. It's given in the problem that the frequency increases, so let's use Doppler logic to get the signs right.

First, let's assume the situation in which the observing sub is stationary. Given the difference in speeds, this means that the sub producing the sound is closing in. The Doppler effect says that the frequency would increase in such a situation. The only way for [tex]\frac{v}{v \mp v_s}[/tex] to increase the frequency is if the denominator is less than the numerator, so we use a minus on the bottom.

For the top, if the sub producing the sound was stationary, the two subs would be drifting apart, meaning that the frequency would decrease. In order for the fraction to decrease the frequency, the numerator must be smaller than the denominator, so we use a minus on the top as well.

So, let's look at the problems:

a. you can use the above formula with the speed of the firetruck as [tex]v_s[/tex] and the speed of the car as [tex]v_o[/tex] to determine the answer.

b. Since the chirps reflect back off the wall, the bat is flying into the reflection. Since the sound and the receiver are coming together, the frequency will increase. A use of the above formula with an arbitrary speed of the bat will show this. However, the bat is also in the same location that the sound is produced so it is a bit more complicated since it is hearing both the reflection and the produced sound. You could say that, although the produced sound did not change frequency, the frequency of the reflection did increase.

c. If both forks are in the same room, the two waves will travel at the same speed because frequency and wavelength are proportional to each other by a constant speed of sound ([tex]v=\lambda \cdot f[/tex]).

Hope that helped.
Steven

Edit: upon rereading your original post, it seems that these questions are over the general concept of the Doppler effect, rather than the mathematics behind it. In that case, let me explain that in the context of the problems:

a. Since the frequency of sound increases when objects come together and decreases when they diverge, whichever of the two objects the firetruck is approaching the fastest will hear the highest frequency. Using reference frames, the firetruck is approaching the driver of the van at 40 m/s and the driver of the car at 10 m/s, so the van will hear a higher frequency.

You were correct for b and c. Sorry for making the problems more complicated than they were.

 

[Mayy]

Thank you, Steven c:

Ahh, I totally had no idea about the formula, because I've never seen it before XD

But as for the rest, thank you so much! It makes so much sense now! (: