1.

In Qunick's acoustic interferometer , it is found that the sound intensity has a minimum value of 100 units at one position of the sliding tube , and cocontinuously climbs to a maximum of 900 units at a second position 1.65 cm from the first . Find (a) the frequency of the sound emitted by the source and (b) the relative amplitudes of the two waves arriving at the detector. Velocity of sound in air= 340 m//s.

Answer» <html><body><p></p>Solution :a. Let `d_(1)` be the length of the <a href="https://interviewquestions.tuteehub.com/tag/fixed-457597" style="font-weight:bold;" target="_blank" title="Click to know more about FIXED">FIXED</a> path and `d_(2)` be the length of other path that can be varied . Then for minimum <a href="https://interviewquestions.tuteehub.com/tag/sound-648690" style="font-weight:bold;" target="_blank" title="Click to know more about SOUND">SOUND</a> ` d_(2) - d_(1)= ( <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> + 1) lambda`. When the adjustable tube is <a href="https://interviewquestions.tuteehub.com/tag/moved-7266899" style="font-weight:bold;" target="_blank" title="Click to know more about MOVED">MOVED</a> out by `x` , the length of the path becomes ` d_(2) + 2 x`. For the next maximum sound <br/> `( d_(2) + 2 x ) - d_(1) = ( n + (1)/( 2)) lambda + ( lambda)/(2) = n lambda+ lambda` <br/> Subtracting ` 2 x= ( lambda)/( 2) orx = ( lambda)/( 4)` <br/> Here `x = 1.65 cm`. <br/> `:. lambda = 4 xx 1.65 = 6.6 cm = 6.6 xx 10^(-2) m` <br/> `C = n lambda or n = ( c)/( lambda) = (340)/( 6.6 xx 10^(-2)) = 5152 Hz` <br/> b. If `a and b`are the amplitudes of the waves , then minimum amplitude ` = a - b` and maximum amplitude ` = a + b` . <br/>` :. (I_(<a href="https://interviewquestions.tuteehub.com/tag/min-548008" style="font-weight:bold;" target="_blank" title="Click to know more about MIN">MIN</a>))/(I_(max)) = (( a - b)^(2))/(( a + b)^(2)) or (100)/(900) = (( a - b)^(2))/(( a + b)^(2))` <br/> or ` (a)/(b) = (2)/(1)`</body></html>


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