Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 30 cm.
a. What is the wavelength of the sound?
b. If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

λ/2 = 10

λ = 20 cm

b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

andromache andromache · Answer 2 · 2022-02-15T16:30:56+01:00

a. The wavelength of the sound = λ = 20 cm

b. The next distance of maximum intensity should be 40 cm

Calculation of the wavelength & next distance:

a. The wavelength should be

Since

(nλ + λ/2) - nλ = 30-20

So,

λ/2 = 10

λ = 20 cm

b. Now the distance should be

The path difference must be increased (n+1) λ.

And, The distance should increase by λ/2 from the point of zero intensity.

So,

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

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