Two loudspeakers, 4.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.35 m. Assume the speed of sound is 340 m/s. A. What is the frequency of the sound? B. If the frequency is then increased while you remain 0.35 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

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AFOKE88 AFOKE88 · Answer 1 · 2020-03-15T02:49:32+01:00

Answer:

A) Frequency of Sound B = 242.85 Hz

B) Frequency at maximum sound intensity = 485.71 Hz

Explanation:

From intensity and path difference equations of sound waves;

the path difference for maximum intensity is x = nλ

Now, for minimum intensity, the path difference(x) is given as;

x = (2n - 1)(λ/2)

Since he has moved 0.35m,the total Path difference = (0.35) + (0.35) = 0.7 m

As n = 1,

0.7 = =(2(1) - 1)(λ/2)

So, (λ/2) = 0.7

And thus λ = 0.7 x 2 = 1.4m

From wavelength equation, we know that; λ = v/f and f = v/λ

Where v is velocity of sound

Thus ; frequency(min) = 340/1.4= 242.85 Hz

B) Now for max the path difference;

From earlier said, x = nλ

So at n=1 and λ = 0.7

x = 1 x 0.7 = 0.7

So following same equation for frequency, fmax = 340/0.7 = 485.71 Hz