A fork gives 5 beats/s with a 40 cm long sonometer wire. If the length of the wire is shortened by 1 cm, the number of beats/s is still the same. The frequency of the fork is

385 Hz

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320 Hz

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395 Hz

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400 Hz

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Solution

The correct option is C 395 Hz
$2 \times b e a t s = δ_{2}^{s} - δ_{1}^{s}$
$2 \times b e a t s = \frac{v}{2 (39)} - \frac{v}{2 (40)}$
$2 \times 5 = \frac{v}{2} (\frac{1}{39} - \frac{1}{40})$
$v = \frac{2 \times 5 \times 2 \times 39 \times 40}{40 - 39}$
$= 31200 c m / s$
$= 312 m / s$
Frequency $(δ_{1}^{s}) = \frac{v}{2 l} = \frac{31200}{2 (40)} = 390 H_{z}$
$B e a t = f r e q u e n c y o f f o r k - f r e q u e n c y (δ_{1}^{s})$
$5 = f r e q u e n c y o f f o r k - 390$
$\Rightarrow$ $F r e q u e n c y o f f o r k$ $= 395 H_{z}$

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A fork gives 5 beats/s with a 40 cm long sonometer wire. If the length of the wire is shortened by 1 cm, the number of beats/s is still the same. The frequency of the fork is

The correct option is C 395 Hz2×beats=δs2−δs12×beats=v2(39)−v2(40)2×5=v2(139−140)v=2×5×2×39×4040−39 =31200 cm/s =312 m/sFrequency(δs1)=v2l=312002(40)=390 HzBeat=frequency of fork−frequency(δs1) 5=frequency of fork−390⇒ Frequency of fork =395Hz