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AB is a cylinder of length 1.0m fitted with a thin flexible diaphragm C at the middle and two others thin portions AC and BC contains hydrogen and oxygen gases , respectively . The diaphragms A and B are set into vibrations of same frequency . What is the minimum frequency of these vibrations for which the diaphragm C is a node ? Under the conditions of the experiment , the velocity of sound in hydrogen is 1100 m//s and in oxygen is 300 m//s. |
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Answer» SOLUTION :When diaphragms `A` and `B` are set in oscillations , ANTINODES are formed at `A` and `B` while a node is formed at `C` ( given) Also `AB = 1.0 m` and ` AC = CB = l (say) = 1//2 = 0.5 m` The portions `AC and BC` behave as closed pipes . In a closed pipe , the modes of vibrations are given by `l =(lambda)/(4) , ( 3lambda)/(4), ( 5 lambda)/(4) , ....` i.e., ` l= ( 2 r + 1) (lambda)/( 4) , r = 0 , 1 , 2 , 3 ,...` or ` lambda = ( 4l)/(2 r + 1) r = 0 , 1 , 2, 3 ,....` Forhydrogen ` lambda_(1) = 4l//( 2 r_(1) + 1)` For oxygen ` lambda_(2) = 4 l//( 2r_(2) + 1)` In both gases the frequency is same As ` n_(1) = n_(2)` or `(v_(1))/( lambda_(1)) = (v_(2))/(lambda_(2))` `(v_(1))/(v_(2)) = (lambda_(1))/( lambda) = (l_(1))/(l_(2)) XX (( 2 r_(2) + 1))/(( 2 r_(1) + 1))` As `l_(1) = l_(2) = 0.5 m` i.e., ` (v_(1))/( v_(2)) =( 2 r_(2) + 1) /( 2 r_(1) + 1)` ` = (1100)/(300) = ( 2r_(2) + 1)/( 2 r_(1) + 1)` i.e., `( 2 r_(1) + 1)/( 2 r_(2) + 1) = (3)/(11)` For minimum frequency , the integer`r_(1) and r_(2)` should be least . Therefore by the inspection ` r_(1) = 1 and r_(2) = 5` Therefore , the frequency of oscillations is given by `n_(min) = (v_(1))/( lambda_(1)) = ( 2 r_(1) + 1) ( v_(1))/( 4 l)` `= ( 2xx 1 + 1) xx ( 1100)/(4 xx 0.5) = ( 3 xx 1100)/(2) = 1650 HZ` i.e., `( 2 r_(1) + 1)/( 2 r_(2) + 1) = (3)/(11)` For minimum frequency , the INTEGERS `r_(1) and r_(2)`should be least. Therefore by inspection `r_(1) = 1and r_(2) = 5` Therefore , the frequency of oscillations is given by `n_(min) = (v_(1))/( lambda_(1)) = ( 2 r_(1) + 1) v_(1)/( 4 l)` ` = ( 2 xx 1 + 1) xx ( 1100)/( 4 xx 0.5) = ( 3 xx 1100)/(2) = 1650 Hz` |
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