A turning fork gives $4$ beats with $50 c m$ length of a sonometer wire. If the length of the wire is shortened by $1 c m$ then no of beats still the same. The frequency of the fork is:

396 H z

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400 H z

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404 H z

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384 H z

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Solution

The correct option is A $396 H z$
Let the frequency of turning fork be $n$

as the frequency of vibration

$s t r i n g α \frac{1}{l e n g t h o f s t r i n g} o r n α \frac{1}{l} ∴ n l = c o n s t a n t o r, n_{1} l_{1} = n_{2} l_{2} (n + 4) 49 = (n - 4) 50 o r, 49 n + 4 \times 49 = 50 n - 200 o r, n = 396$

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A turning fork gives 4 beats with 50 cm length of a sonometer wire. If the length of the wire is shortened by 1 cm then no of beats still the same. The frequency of the fork is:

The correct option is A 396HzLet the frequency of turning fork be n as the frequency of vibration stringα1lengthofstringornα1l∴nl=constantor,n1l1=n2l2(n+4)49=(n−4)50or,49n+4×49=50n−200or,n=396

A turning fork gives $4$ beats with $50 c m$ length of a sonometer wire. If the length of the wire is shortened by $1 c m$ then no of beats still the same. The frequency of the fork is: