Question

Two loudspeakers, 4.0 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 0.25 m.
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answer 1

Answer:

a) 343.0 Hz b) 686.0  Hz

Explanation:

a) First, we need to know the distance to both speakers.

If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.

So, if  he moves 0.25 m towards one of them, the distance from any speaker will be as follows:

d₁ = 2.0 m-0.25 m= 1.75 m

d₂ = 2.0 m + 0.25 m = 2.25 m

The difference between these distances is the path difference between the sound from both speakers:

d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m

If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:

d = (2*n-1)*(λ/2) = 0.5 m

The minimum distance is for n=1:

⇒ λ = 2* 0.5 m = 1 m

In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:

v = λ*f, where v= 343 m/s and λ=1 m.

Solving for f, we have:

$f =\frac{343.0 m/s}{1.0 m} = 343 Hz$

b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met  the following condition:

d = 0.5 m = 2n*(λ/2) = λ (for n=1)

if the speed remains the same (343 m/s) we can find the new frequency as follows:

$f =\frac{v}{d} =\frac{343 m/s}{0.5m} =686.0 Hz$

⇒ f = 686.0 Hz