Question

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? The speed of sound in air = 320 m/s.

Answer 1

Answer ⇒ Frequencies are 1200 Hz,  2000 Hz,  2800 Hz,  3600 Hz, 4400 Hz.

Explanation ⇒ We know that the destructive interference will be possible only when the phase difference between the two waves is having odd multiple of π.  

Now, As per as the question,

the path difference = 6.4 - 6.0

= 0.40 m

Thus, the phase difference  

Φ = 2π × 0.40/

∴ Φ = 2π × 0.40 × ν/V

∴ Φ = 2π× 0.40 × ν/320

∴ Ф = π × (0.0025 ν)

Now, for the Ф to be an odd multiple of π, 0.0025ν must be an odd number,

Therefore,

0.0025ν = 1, 3, 5, 7, 9, 11, 13, .....

ν/400 = 1, 3, 5, 7, 9, 11, 13, .....

ν = 400 Hz,  1200 Hz,  2000 Hz,  2800 Hz, 3600 Hz,  4400 Hz,.........

But the range of the given frequency is in between 500 Hz to 5000 Hz.

Hence the destructive interference at the place for the given frequency range is for the frequencies 1200 Hz,  2000 Hz,  2800 Hz,  3600 Hz, 4400 Hz.

Hope it helps.

Answer 2

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