Question

A wire having tension $225 \, N$ produces six beats per second when it vibrates with some tuning fork. When the tension changes to $256 \, N$ , it vibrates with the same fork with the number of beats remaining unchanged. The frequency of the fork will be

• A. $186 \, Hz$
• B. $225Hz$
• C. $256Hz$
• D. $280Hz$

Answer 1

Answer: (A)

For wire vibrating under tension, the fundamental frequency is given by
$f_{1}=\frac{1}{2 L}\sqrt[ \, ]{\frac{T}{\mu }}$ , Where T is tension and $\mu $ linear mass density of string
It is given that, when it is sounded with a tuning fork of frequency $x$ , $6$ beats per second were heard
$\therefore \, \left(x - f\right)=6$
$f_{1}=\frac{1}{2 L}\sqrt{\frac{225}{\mu }}$
$f_{2}=\frac{1}{2 L}\sqrt{\frac{256}{\mu }}$ undefined
$\therefore \, \left(x + 6\right)=\frac{16}{15}\left(\right.x-6\left.\right)$
$\therefore \, 15x+90=16x-96$
$\therefore \, x=186 \, Hz$