Question

A tuning fork of frequency $392\, Hz$, resonates with $50\, cm$ 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
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Where $l$ is length of string,
$T$ is tension and $m$ is mass per unit length of string.
Given, $ l_{1}=50\, cm ,$
$ l_{2}=\left(1-\frac{2}{100}\right) \times 50$
$=49 \,cm$
$\frac{n_{1}}{n_{2}}=\frac{l_{2}}{l_{1}}=\frac{49}{50}$
$\Rightarrow n_{2}=\frac{50}{49} \times 392$
$=400 \,Hz$
Also number of beats per second (beat frequency)
$=n_{2}-n_{1}$
$=$ difference of the frequencies of sound-sources.
$=400-392=8$