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

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14.Waves and Sound

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A tuning 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

(c) $n \propto \frac{1}{l}$ ==> $\frac{{\Delta n}}{n} = – \frac{{\Delta l}}{l}$

If length is decreased by $2\%$ then frequency increases by $2\%$ i.e., $\frac{{{n_2} – {n_1}}}{{{n_1}}} = \frac{2}{{100}}$

==> ${n_2} – {n_1} = \frac{2}{{100}} \times {n_1} = \frac{2}{{100}} \times 392 = 7.8 \approx 8.$

Standard 11

Physics

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