Question

Two loudspeakers are placed on a wall 3.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference between the two waves when they reach the observer? (Your answer should be between 0 and 2.) rad (b) What if? What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

Answer 1

Answer:

Part a)

$\Delta \phi = 2.2 \pi$

Part b)

$f = 411.3 Hz$

Explanation:

As we know that the observer is standing in front of one speaker

So here the path difference of the two sound waves reaching to the observer is given as

$\Delta x = 3\sqrt2 - 3$

$\Delta x = 1.24 m$

now phase difference is related with path difference as

$\Delta \phi = \frac{2\pi}{\lambda}(\Delta x)$

$\Delta \phi = \frac{2\pi}{\lambda}(1.24)$

here in order to find the wavelength

$\lambda = \frac{c}{f}$

$\lambda = \frac{340}{300} = 1.13$

now we have

$\Delta \phi = \frac{2\pi}{1.13}(1.24) = 2.2\pi$

Part b)

Now we know that when phase difference is odd multiple of $\pi$

then in that case the the sound must be minimum

So nearest value for minimum intensity would be

$\Delta \phi = 3\pi$

so we have

$3\pi = \frac{2\pi}{\lambda}(1.24)$

so we have

$\lambda = 0.827$

now we have

$\frac{340}{f} = 0.827$

$f = 411.3 Hz$