A vertical smoked plate falls vertically under gravity and traces found by transverse vibrations of a fork show $10$ complete vibrations in each of two consecutive lengths which differ by $1.8 c m$ . The frequency of the fork is

233.3 H z

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333.3 H z

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444.3 H z

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555.3 H z

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Solution

The correct option is B $233.3 H z$
Let the length of $10$ waves in two sets be $l_{1}$ and $l_{2}$ .
Then, $l_{1} = u t + \frac{1}{2} g t^{2}$ and $l_{2} = u_{1} t + \frac{1}{2} g t^{2}$
But $u_{1} = u + g t$

Hence, $l_{2} - l_{1} = g t^{2}$ or $t = \sqrt{\frac{l_{2} - l_{1}}{g}}$

Now frequency of the fork is given by:
$ν = N / t = 10 \times \sqrt{\frac{g}{l_{2} - l_{1}}} = 10 \times \sqrt{\frac{9.8}{0.018}} = 233.3 H z$

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A vertical smoked plate falls vertically under gravity and traces found by transverse vibrations of a fork show 10 complete vibrations in each of two consecutive lengths which differ by 1.8cm. The frequency of the fork is

The correct option is B 233.3HzLet the length of 10 waves in two sets be l1 and l2.Then, l1=ut+12gt2 and l2=u1t+12gt2 But u1=u+gtHence, l2−l1=gt2 or t=√l2−l1gNow frequency of the fork is given by:ν=N/t=10×√gl2−l1=10×√9.80.018=233.3 Hz

A vertical smoked plate falls vertically under gravity and traces found by transverse vibrations of a fork show $10$ complete vibrations in each of two consecutive lengths which differ by $1.8 c m$ . The frequency of the fork is