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In Kundt's tube experiment the following observations were made : Length ofthe brass rod is 100 cm, average length of a loop in air is 10.3 cm and in carbon-di-oxide=8.0 cm. Calculate thevelocityofsound in brass and in CO^(2). Whatisthefrequency ofthe note ? |
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Answer» Solution :LET `lambda_(R), lambda_(a)` and`lambda_(g)` be the wavelengths in cm in therod, in air and in gas respectively. Then `(lambda_(r))/(2) = 100 or lambda_(r) = 200 cm` `(lambda_(a))/(2) = 10.3 `or `lambda_(a) = 20.6 cm ``(lambda_(g))/(2) = 8.0 or lambda_(g) = 16cm` Let `v_(r), v_(a)` and `(v_(g))` be the sound velocities in rod, air and gas respectively.Then `(v_(r))/(v_a) = (lambda_(r))/(lambda_a) or v_(r) = (lambda_(r))/(lambda_(a)) xx v_(a)` `v_(r) = (200)/(20.6) xx 350` = `3.4 xx 10^(3) m//s` `v_(g) = 350 xx (16)/(20.6) ` = `271.8 m//s` If n be frequency of the note, then we have `V'_(a) = nlambda_(a)` `n = (v_(a))/(lambda_(a)) = (350)/(0.206)` = 1699.02 HZ |
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