Question

A wire produces a fundamental frequency $n$ under a tension $T$. The wire is stretched to twice its length and kept under the same tension. The fundamental frequency of the wire is

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

Answer 1

The correct option is D $\frac{n}{\sqrt{2}}$

$f=\frac{\nu }{2l}=\frac{\sqrt{\frac{T}{m/l}}}{2l}=\frac{1}{2}\sqrt{\frac{T}{ml}}=n$
when wire is stretched to twice the length $l$ becomes $2l$.
So, $f=\frac{1}{2}\sqrt{\frac{T}{2ml}}=\frac{1}{\sqrt{2}}\times \left(\frac{1}{2}\sqrt{\frac{T}{ml}}\right)$
$=\frac{1}{\sqrt{2}}\times n$
new frequency $=\frac{n}{\sqrt{2}}$