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Two speakers are driven by the same oscillator with frequency of 200 Hz. They are located 4 m part on a vertical pole. A man walks straight towards the lower speaker in a direction perpendicular to the pole, as shown in figure. (i) How many times will he hear a minimum in sound intensity, and (ii) how far is he from the pole and at these moments ? Take the speed of sound to be 330 m/s, and ignore any sound reflections coming off the ground.
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Text Solution

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The correct Answer is:
(i) 2
(ii) 9.28 m and 1.99 m
(i) `lamda=(v)/(f)=(330)/(200)=1.65m,(d)/(lamda)=(4)/(1.65)=2.4242`
At infinity, path difference =0
As the approaches, the path difference changes as
`0,(lamda)/(2),lamda,(3 lamda)/(2),2 lamda`
`:.` Hence only minima will appear to man
(ii) For `Delta=(lamda)/(2)=sqrt(d^(2)+x_(1)^(2))-x_(1)and (3lamda)/(2)=sqrt(d^(2)+x_(2)^(2))-x_(2)`
`rArr x_(1)=9.28m,x_(2)=1.99 m`
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