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Question

An oscillator of frequency 680 Hz drives two speakers. The speakers are fixed on a vertical pole at a distance 3 m from each other as shown in figure. A person whose height is almost the same as that of the lower speaker walks towards the lower speaker in a direction perpendicular to the pole. Assuming that there is no reflection of sound from the ground and sound speed is v=340 m/s, choose the correct option(s):


A
The person hears the first minimum sound intensity at 18 m away from the pole.
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B
The person hears the second minimum intensity at a distance of 5.6 m from the speaker.
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C
The person hears a minimum sound intensity for a total of 6 times.
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D
The person hears a maximum sound intensity for a total of 8 times.
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Solution

The correct options are
A The person hears the first minimum sound intensity at 18 m away from the pole.
B The person hears the second minimum intensity at a distance of 5.6 m from the speaker.
C The person hears a minimum sound intensity for a total of 6 times.
Frequency of source = 680 Hz
and v=340 m/s
From formula
wavelength (λ)=vf=340680=0.5 m
Let the person be at distance d when the observed first minimum intensity.
Hence, path difference (Δx)=λ2
S2AS1A=λ2
d2+9d=λ2=14
d2+9=d+14
d=17.8717.9 m


For second time,
Destructive interference occurs at 32λ
So
Δx=32λ
d2+9d=32×12
d2+9=34+d
d=5.6 m

For maximum (i.e constructive interference)
path difference =nλ
If final maximum is obtained at foot of the pole,
Distance between the sources =3=nλ
3=n(12)
n=6
Therefore, total 6 maximum and 6 minimum intensity will be formed.
Option (a), (b), (c) are correct.

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