Two tuning forks A and B vibrating simultaneously produce $5$ beats. Frequency of B is $512$ $H_{z}$ . It is seen that if A is filed a little, then the number of beats occur at shorter interval. Frequency of A in $H_{z}$ will be.

502

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507

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517

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522

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Solution

The correct option is C $517$
$n_{1} = x$ $n_{2} = x - 5$
$n_{1} l_{1} = n_{2} l_{2}$
$n_{2} = 512 H z$
So $512 = x - 5$
$x = 512 + 5$
$H_{1} = n = 517 H z$

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Two tuning forks A and B vibrating simultaneously produce 5 beats. Frequency of B is 512 Hz. It is seen that if A is filed a little, then the number of beats occur at shorter interval. Frequency of A in Hz will be.

The correct option is C 517n1=x n2=x−5n1l1=n2l2n2=512 HzSo 512=x−5x=512+5H1=n=517 Hz

Two tuning forks A and B vibrating simultaneously produce $5$ beats. Frequency of B is $512$ $H_{z}$ . It is seen that if A is filed a little, then the number of beats occur at shorter interval. Frequency of A in $H_{z}$ will be.

The correct option is C $517$
$n_{1} = x$ $n_{2} = x - 5$
$n_{1} l_{1} = n_{2} l_{2}$
$n_{2} = 512 H z$
So $512 = x - 5$
$x = 512 + 5$
$H_{1} = n = 517 H z$