Question

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

Answer 1

The correct option is C 8

(c) $n\propto \frac{1}{l}\Rightarrow \frac{\Delta n}{n}=-\frac{\Delta l}{l}$If length is decreased by $2\%$ then frequency increases by $2\%$

i.e, $\frac{n_2-n_1}{n_1}=\frac{2}{100}$

$\Rightarrow n_2-n_1=\frac{2}{100}\times n_1=\frac{2}{100}\times 392=7.8\approx 8$