Question

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

Answer: (C)

According to law of length $n\propto \frac{1}{l}$.
The frequency of tuning fork is given by
$n=\frac{1}{2l}\sqrt{\frac{T}{m}}$
Where $l$ is length of string,
$T$ is tension and $m$ is mass per unit length of string.
Given, $l_1=50cm,$
$l_2=\left(1-\frac{2}{100}\right)\times 50$
$=49cm$
$\frac{n_1}{n_2}=\frac{l_2}{l_1}=\frac{49}{50}$
$\Rightarrow n_2=\frac{50}{49}\times 392$
$=400Hz$
Also number of beats per second (beat frequency)
$=n_2-n_1$
$=$ difference of the frequencies of sound-sources.
$=400-392=8$