# Sound waves . . . help!

1. Mar 2, 2006

### motherlovebone

I have been stumped on these problems for about a half an hour now, and I need some big help on them!

Problem 1:
"A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. An echo is heard 2.60 s later. How far away is the reflecting object?"

I guessed that 2.60 s was the period, so I found the reciprocal to get the frequency. Once I did that, I put the speed of sound in for v in the equation v=frequency x wavelength. My answer, 891.8 m, sounds preposterous however.

Problem 2:
"The notes produced by a violin range in frequency from approximately 196 Hz to 2637 Hz. Find the possible range of wavelengths produced by the instrument when the speed of sound is 340 m/s."

For this one, would I use v=frequency x wavelength? I did 340 divided by 196, which was 1.735, then 340 divided by 2637, which was 0.129. So would the range of wavelengths be 0.129 to 1.735?

Last edited: Mar 2, 2006
2. Mar 2, 2006

### LeonhardEuler

You know the distance, "d" traveled by the wave is equal to the velocity times the time: $$d=vt$$. So if the sound is heard 2.6s later then the distance it must have traveled would be the distance from the boat to the wall plus the distance from the wall to the boat. In other words, twilce the distance fom the wall.
The wavelength, $\lambda$ is equal to to the speed of the wave, v, over the frequency, f:
$$\lambda=\frac{v}{f}$$
So the maximum and minimum frequency give you the minimum and maxximum wavelength. By the way, in case your interested in where this formula came from, it makes sense. A frequency of, say, 2 Hz means that 2 waves pass you in 1s. If the waves are traveling at 10 meters every second and two pass you in a second, then they each must be 5 meters long, so $\lambda=\frac{v}{f}$