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18 votes
Two loudspeakers placed 6.00 m apart are driven in phase by an audio oscillator having a frequency range from 1908 Hz to 2471 Hz. A point P is located 4.70 m from one loudspeaker and 3.60 m from the other speaker. At what frequency of the oscillator does the sound reaching point P interfere constructively

User Baziorek
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1 Answer

6 votes
6 votes

Answer:

2164 Hz

Step-by-step explanation:

Since point P is 4.70 m away from one speaker and 3.60 m away from the other speaker, the path length difference ΔL = 4.70 m - 3.60 m = 1.1 m.

The path length difference ΔL = nλ for a constructive interference where n is an integer and λ = wavelength of sound from oscillator = v/f where v = speed of sound in air = 340 m/s and f = frequency of sound from oscillator.

So, ΔL = nλ = nv/f

So, the frequency from the oscillator is f = nv/ΔL

Substituting the values of v and ΔL into the equation, we have

f = nv/ΔL

f = n340 m/s/1.1 m

f = n309.09 /s

f = 309.09n Hz

We now insert values of n that will gives us a frequency in the range 1908 Hz to 2471 Hz.

The value of n that will give us a frequency in the range is n = 7

So, when n = 7,

f = 309.09n Hz

f = 309.091 × 7 Hz

f = 2163.64 Hz

f ≅ 2164 Hz

So, the frequency of the oscillator that will produce a constructive interference at P is 2164 Hz.

User Lennin
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