When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 c m$ or $100 c m$ . Then the frequency of the tuning fork is (consider that the tension in the wire is kept constant):

1616 H z

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1608 H z

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1632 H z

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1600 H z

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Solution

The correct option is C $1608 H z$
We know that: $n l = c o n s t a n t$
$\Rightarrow n_{1} l_{1} = n_{2} l_{2}$
$\Rightarrow (n + 8) \times 100 = (n - 8) \times 101$
$\Rightarrow \frac{n + 8}{n - 8} = \frac{101}{100}$
$\Rightarrow \frac{n}{8} = 201$
$\Rightarrow n = 1608 H z$

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When a vibrating tuning fork is placed on a sound box of a sonometer, 8 beats per second are heard when the length of the sonometer wire is kept at 101 cm or 100 cm. Then the frequency of the tuning fork is (consider that the tension in the wire is kept constant):

The correct option is C 1608 HzWe know that: nl=constant⇒n1l1=n2l2⇒(n+8)×100=(n−8)×101⇒n+8n−8=101100⇒n8=201⇒n=1608 Hz

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 c m$ or $100 c m$ . Then the frequency of the tuning fork is (consider that the tension in the wire is kept constant):

The correct option is C $1608 H z$
We know that: $n l = c o n s t a n t$
$\Rightarrow n_{1} l_{1} = n_{2} l_{2}$
$\Rightarrow (n + 8) \times 100 = (n - 8) \times 101$
$\Rightarrow \frac{n + 8}{n - 8} = \frac{101}{100}$
$\Rightarrow \frac{n}{8} = 201$
$\Rightarrow n = 1608 H z$