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?

Question

contestada

Respuesta :

aristocles aristocles · Answer 1 · 2019-08-23T03:13:43+02:00

Answer:

Part a)

[tex]\Delta \phi = 2.2 \pi[/tex]

Part b)

[tex]f = 411.3 Hz[/tex]

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

[tex]\Delta x = 3\sqrt2 - 3[/tex]

[tex]\Delta x = 1.24 m[/tex]

now phase difference is related with path difference as

[tex]\Delta \phi = \frac{2\pi}{\lambda}(\Delta x)[/tex]

[tex]\Delta \phi = \frac{2\pi}{\lambda}(1.24)[/tex]

here in order to find the wavelength

[tex]\lambda = \frac{c}{f}[/tex]

[tex]\lambda = \frac{340}{300} = 1.13[/tex]

now we have

[tex]\Delta \phi = \frac{2\pi}{1.13}(1.24) = 2.2\pi[/tex]

Part b)

Now we know that when phase difference is odd multiple of [tex]\pi[/tex]

then in that case the the sound must be minimum

So nearest value for minimum intensity would be

[tex]\Delta \phi = 3\pi[/tex]

so we have

[tex]3\pi = \frac{2\pi}{\lambda}(1.24)[/tex]

so we have

[tex]\lambda = 0.827 [/tex]

now we have

[tex]\frac{340}{f} = 0.827[/tex]

[tex]f = 411.3 Hz[/tex]