# Frequency of first intensity maximum for loudspeakers

1. Dec 13, 2008

### vaizard

1. The problem statement, all variables and given/known data
Two loudspeakers are set up 4.0 meters apart and are driven in phase by the same amplifier. The listener is located a sufficient distance away so the angle to the listener is approximately the same at each speaker as indicated in the drawing. The speed of sound is 344 m/s. What frequency would the speakers need to emit if the first intensity maximum from the central maximum occurs at an angle of 30° from the perpendicular to each speaker?

2. Relevant equations
$$v=\lambda \cdot f$$
There is also $$A = 2 A_0 cos(\frac{\phi}{2})$$ in the solution to the problem but I don't understand where this equation even came from. It's not in my book or anything.

3. The attempt at a solution
Well, I know I'm looking for f, the frequency. So I did $$f = \frac{v_{\mbox{sound}}}{\lambda}$$ but that's about it. Now the actual solution uses $$A=2A_0 cos(\frac{\phi}{2})$$ and $$\phi = k \Delta x$$. I would like to understand where these equations came from. Are they universal for this type of problem, or were they somehow derived from the given information?

Here is a copy of the whole problem with the solution.

Last edited: Dec 13, 2008
2. Dec 15, 2008

### Carid

Surely the first max will occur when the path length from one speaker is a single wavelength longer than to the path length to the other.
The path difference shown here is 4*sin30° = 2metres
So 344m/sec = 2m * frequency
Frequency = 172 Hertz.

Last edited: Dec 15, 2008