Lowest Possible Frequencies of a Sound

[nautikal]

Homework Statement

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?

Homework Equations

$$\lambda=v/f$$

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).

 

[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

I know that there is some maximum wavelength where there will be crests at a distance of 40m and 30m...

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.

 

[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 don't really know what you would have to do that.

 

[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 :)

 

[rwisz]

I would approach this problem by first understanding the relationship between wavelength, frequency, and the speed of sound. The equation \lambda=v/f represents the relationship between wavelength (\lambda), frequency (f), and the speed of sound (v). This equation tells us that as the frequency of a sound increases, the wavelength decreases, and vice versa.

In this problem, we are given the speed of sound (v=340 m/s) and the distances at which the crests of the sound wave are passing by listeners (x=40m and y=30m). From this information, we can set up the following equations:

For the listener at (40,0):
\lambda = 40m
For the listener at (0,30):
\lambda = 30m

We can then use the equation \lambda=v/f to solve for the lowest possible frequencies for each listener. Plugging in the given values for wavelength and speed of sound, we get:

For the listener at (40,0):
40m = (340 m/s)/f
f = 340 m/s / 40m = 8.5 Hz

For the listener at (0,30):
30m = (340 m/s)/f
f = 340 m/s / 30m = 11.3 Hz

Therefore, the lowest two possible frequencies for the sound in this scenario are 8.5 Hz and 11.3 Hz. It's important to note that these are not the only possible frequencies, but they are the lowest possible ones based on the given information.