# Lowest Possible Frequencies of a Sound

1. Feb 23, 2009

### nautikal

1. The problem statement, all variables and given/known data
A loudspeaker at the origin emits sound waves on a day when the speed of sound is 340 m/s. A crest of the wave simultaneously passes listeners at the (x, y) coordinates (40m, 0) and (0, 30m). What are the lowest two possible frequencies of the sound?

2. Relevant equations
$$\lambda=v/f$$

3. The attempt at a solution
I know the answer is 34 and 68 Hz from looking in the back of the book, but I don't understand how to get there. I know that there is some maximum wavelength where there will be crests at a distance of 40m and 30m, but I don't know how to solve for that mathematically (it is 10m by guess and check using common factors).

2. Feb 23, 2009

### S_Happens

It's pretty straight forward. You have one equation with three variables. To solve for frequency (one variable) you must have the two other variables. One is given directly to you, and the other is found with simple arithmetic.

You don't really care that that the crests are simultaneously passing the locations 30m and 40m from the source. What matters is what that statement implies. Hint: The source location doesn't matter.

You made the connection when you said
What commonality does 10m have with two positions of 30m and 40m?

Anything more and I'm giving you the answer you already didn't work for.

3. Feb 19, 2010

### Epa06

So 10m is the wavelength, and when you plug it into the lambda=v/f and solve for f, you get 34 Hz. But how do you get the 68 Hz? I know to get 68 Hz, you divide 340 by 5, but i dont really know what you would have to do that.

Last edited: Feb 19, 2010
4. May 17, 2011

### Smazmbazm

Think about the question...it's asking for the two lowest frequencies. Fundamental and then the Second Harmonic. 34Hz is the Fundamental (First Harmonic) so the Second Harmonic being double the Fundamental is...68Hz :)