Given:Moe, Larry, and Curly stand in a line with a spacing of d = 1.60 m. Larry is 3.00 m in front of a pair of stereo speakers 0.800 m apart, as shown in the figure . The speakers produce a single-frequency tone, vibrating in phase with each other.

Question: What are the two lowest frequencies that allow Larry to hear a loud tone while Moe and Curly hear very little?
Answer in kHz.

Okay I don't where to start. A little help?

It's just like the experiment where you have a double slit interference.

Use the formula \lambda = \frac{dx}{l}

Where lambda is the wavelength, d the separation between the two sources, x the separation of the 'bright' fringes and l the distance from the source to the three people.

d = 0.800 m
l = 3.00 m
x = 3.2 m and 1.07 m

To get the values of x, you'll perhaps better make a sketch.

Can you understand why is x 3.2 and 1.07 m?

http://p1cture.me/images/77591923647672467276.png

http://p1cture.me/images/93043777809919089402.png

NOTE: Dark represent areas where there is constructive interference, and I did only for the middle and lower part.

OK I understand why x should be 1.6 and 1.07 but I don't understand how you go those numbers?

*got

No, x is 3.2 and 1.07 m, not 1.6 and 1.07 m.

In the first case, you need to have Larry in such a way that he is in a region of constructive interference (dark part) and Moe directly in the next region of destructive interference. This means that half the separation between two regions of constructive interference must be 1.6 m.

The total separation will be 1.6 m x 2 = 3.2 m.

In the second case, Larry should be still in the region of constructive interference, hence the dark region. However, since we already got the first order (in case 1), we look for the second order, and Moe should be in the second next region of destructive interference.

The separation of Moe and Larry is 1 dark region and a half.
So, \frac32x = 1.6\ m

x = 1.07\ m

Sorry I am total idiot haha. Thanks for your help!

No, it's good that you ask. Be sure to understand, if not, I'll try a different approach :)