A tuning of fork of frequency 392, Hz , resonates with 50, cm length of a string under tension (T) .

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14.Waves and Sound

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A tuning of fork of frequency $392\, Hz$, resonates with $50\, cm$ length of a string under tension $(T)$. If length of the string is decreased by $2\%$, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is

A

$4$

B

$6$

C

$8$

D

$12$

Solution

$\mathrm{n} \propto \frac{1}{\ell} \Rightarrow \frac{\Delta \mathrm{n}}{\mathrm{n}}=-\frac{\Delta \ell}{\ell}$

If length is decreased by $2 \%$ then frequency

increases by $2 \%,$ i.e., $\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{n}_{1}}=\frac{2}{100}$

$\Rightarrow \mathrm{n}_{2}-\mathrm{n}_{1}=\frac{2}{100} \times \mathrm{n}_{1}=\frac{2}{100} \times 392=7.8 \approx 8$

Standard 11

Physics

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