Question

A string under tension of $129.6N$ produces $10$ beats/ second, when it vibrates along with a tuning fork. When the tension in the string is increased to $160N$, it vibrates in unison with the tuning fork. Then, frequency of the tuning fork is

• A. $100Hz$
• B. $110Hz$
• C. $90Hz$
• D. $220Hz$
• E. $95Hz$

Answer 1

The correct option is A $100Hz$

Given, $T_1=126.6N$
$T_2=160N$
and $b=10beats/sec$
As, in first medium,
$f_1=\frac{1}{2l}\sqrt{\frac{T_1}{m}}=f_0-b....\left(i\right)$
In the second condition,
$f_2=\frac{1}{2l}\sqrt{\frac{T_2}{m}}=f_0.....\left(ii\right)$
On dividing Eq. (i) by Eq. (ii), we get
$\sqrt{\frac{T_1}{T_2}}=\frac{f_0-b}{f_0}\Rightarrow \sqrt{\frac{129.6}{160}}=\frac{f_0-10}{f_0}$
Frequency of the tuning fork,
$\Rightarrow f_0=100Hz$.