Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is
(a) 2 sec
(b) 0.5 sec
(c) 250 sec
(d) 252 sec

Question

Arun · Answer

n 1 ​ = 2 5 6 n 2 ​ = 2 5 8 B o t h t u n i n g f o r k s a r e m a k i n g b e a t s s o , w e k n o w t h a t t ( f o r m a x . ) ​ = n 2 ​ − n 1 ​ 1 ​ sec = 2 5 8 − 2 5 6 1 ​ = 0 . 5 sec ​ Hence, the option B is the correct answer.

Vikas TU · Answer

T = 1/n1-n2 = 1/258-256 = 1/2 ….............................................................................................................

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Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is (a) 2 sec (b) 0.5 sec (c) 250 sec (d) 252 sec

Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is
(a) 2 sec
(b) 0.5 sec
(c) 250 sec
(d) 252 sec

2 Answers

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Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is (a) 2 sec (b) 0.5 sec (c) 250 sec (d) 252 sec

Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is (a) 2 sec (b) 0.5 sec (c) 250 sec (d) 252 sec

2 Answers

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Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is
(a) 2 sec
(b) 0.5 sec
(c) 250 sec
(d) 252 sec