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!!!
