Question

If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

• A. $\frac{n}{4}$
• B. $\sqrt{2}n$
• C. n
• D. $\frac{n}{\sqrt{2}}$

Answer 1

The Correct Option is C

$n=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2 \rho}}$
$ \rho'_1=\frac{ \rho}{2};T'=2T $ and $ D'=2D$ or $r'=2r$
$n'=\frac{1}{2l}\sqrt{\frac{2T}{\pi(2r)^2\frac{\rho}{2}}}$ after solving, $n'=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2l}}=n$.