Question

Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator, whose frequency range is 1595 to 2158 Hz. A point P is located 4.7 m from one loudspeaker and 3.6 m from the other. The speed of sound is 344 m/s. The frequency produced by the oscillator, for which destructive interference occurs at point P, in SI units is what? SHOW ALL WORK PLEASE! It is either 2001, 2033, 2127, 2095, or 2064 Hz.

Answer 1

Answer:

The frequency produced by the oscillator, for which destructive interference occurs at point P, in the range of frequencies for the oscillator = 2033 Hz

Explanation:

For destructive interference, the difference in path length of the waves from each loudspeaker is related to the wavelength through.

(m + ½)λ = |d₁ - d₂|

where m can take on positive whole number values 0,1,2,3...

Point P is 4.7 m from A and 3.6 m from B

d₁ = 4.7 m

d₂ = 3.6 m

|d₁ - d₂| = 4.7 - 3.6 = 1.1 m

(m + ½)λ = |d₁ - d₂| = 1.1

(m + ½)λ = 1.1

where m could be any whole number from 0,1,2...

And the relationship between velocity of a wave, v, its frequency, f, and the wavelength, λ, is given as

v = fλ

The frequency range of the audio oscillator is frequency range is 1595 to 2158 Hz.

We can find a wavelength for the sound within this range, so as to obtain the exact frequency.

The options include 2001, 2033, 2127, 2095, or 2064 Hz.

Taking just 1 frequency in that range, f = 2033 Hz.

v = fλ

λ = (v/f) = (344/2033) = 0.169 m

Inserting in the destructive interference equation

(m + ½)λ = 1.1

If λ = 0.169 m

(m + ½) = (1.1/0.169) = 6.5

m + ½ = 6.5

Hence, it is evident that m = 6 for this question.

And the corresponding frequency at this level of destructive interference is 2033 Hz

Hope this Helps!!!