A classical mechanics problem by A Former Brilliant Member

Classical Mechanics Level 3

If i have 3 prongs having frequency v , v − 1 , v + 1 v , v-1 , v+1 , find the maximum number of beats that can be heard.

5 Scenario not possible 4 No beat heard due to super-imposition 12 2 3

A Former Brilliant Member by A Former Brilliant Member

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1 solution

it was asked in previously in mains

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Y ( x , t ) = Y(x,t)= A s i n ( k x − w 1 t ) + A s i n ( k x − w 2 t ) + A s i n ( k x − w 3 t ) Asin(kx-w1t)+Asin(kx-w2t)+Asin(kx-w3t) .Where w 1 = 2 π ( n − 1 ) , w 2 = 2 π ( n + 1 ) , w 3 = 2 π n w1=2π(n-1),w2=2π(n+1),w3=2πn .Solving we get Y ( x , t ) = 3 s i n ( k x − 2 π n t ) Y(x,t)=3sin(kx-2πnt) .So I guess no beats can be heard due to superimposition .If the prongs are identical then we can safely assume same A A

Spandan Senapati - 4 years, 2 months ago

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yes A is same and ur final expression is not correct bhai , it 'll be sin 2 sin xt[2cospi t 1] so beat is 2*pi/pi = 2 ! ans

A Former Brilliant Member - 4 years, 2 months ago

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Oh sorry.Calc mistake got it.

Spandan Senapati - 4 years, 2 months ago

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