Question

$5$ beats/second are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95m$ or $1m$. The frequency of the fork will be :

• A. 195 Hz
• B. 150 Hz
• C. 300 Hz
• D. 251 Hz

Answer 1

Answer: (A)

Length of wire is $L_1=0.95m;L_2=1m$.
Number of beats per second heard $=5$.
Let frequency of fork be $f$. Therefore,
$\frac{v}{2L_1}-f=5$
$\Rightarrow \frac{v}{2L_1}=5+f$
and $f-\frac{v}{2L_2}=5$
$\Rightarrow \frac{v}{2L_2}=f-5$
Dividing Eq. (1) by Eq. (2), we get
$\frac{\frac{v}{2L_1}}{\frac{\nu }{2L_2}}=\frac{5+f}{f-5}$
$\Rightarrow \frac{L_2}{L_1}=\frac{f+5}{f-5}$
$\Rightarrow \frac{1}{0.95}=\frac{f+5}{f-5}$
$\Rightarrow f-5=0.95f+4.75$
$\Rightarrow f-0.95f=5+4.75$
$\Rightarrow 0.05f=9.75$
$\Rightarrow f=\frac{9.75}{0.05}$
$\Rightarrow f=195Hz$