A tuning fork produces 4 beats/s with a sonometer wire when its lengths are 50 cm, 51cm. The frequency of that tuning fork is

400 Hz

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404 Hz

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408 Hz

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412 Hz

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Solution

The correct option is A 404 Hz
$f = \frac{v}{2 l}$
$∴ f_{1} = \frac{v}{2 l_{1}}; f_{2} = \frac{v}{2 l_{2}}$
$f_{1} - f_{2} = 8$
$\frac{v}{2} (\frac{1}{l_{1}} - \frac{1}{l_{2}}) = 8$
$\frac{v}{2} (\frac{1}{50 \times 51}) = 8$
$\Rightarrow \frac{v}{2} = 204$
$\Rightarrow v = 408 H z$
As it produces $4$ beats/sec
Frequency, $f = (408 - 4) = 404 H z$

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A tuning fork produces 4 beats/s with a sonometer wire when its lengths are 50 cm, 51cm. The frequency of that tuning fork is

The correct option is A 404 Hzf=v2l∴f1=v2l1;f2=v2l2f1−f2=8v2(1l1−1l2)=8v2(150×51)=8⇒v2=204⇒v=408 HzAs it produces 4 beats/secFrequency, f=(408−4)=404 Hz