Question

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

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

Answer 1

Answer: (A)

For wire vibrating under tension, the fundamental frequency is given by
$f_1=\frac{1}{2L}\sqrt[]{\frac{T}{\mu }}$ , 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
$\therefore \left(x-f\right)=6$
$f_1=\frac{1}{2L}\sqrt{\frac{225}{\mu }}$
$f_2=\frac{1}{2L}\sqrt{\frac{256}{\mu }}$ undefined
$\therefore \left(x+6\right)=\frac{16}{15}(x-6)$
$\therefore 15x+90=16x-96$
$\therefore x=186Hz$