Question

A wire has frequency $f$. Its length is doubled by stretching. Its frequency now will be

• A. 1.4 f
• B. 0.7 f
• C. 2 f
• D. f

Answer 1

Answer: (B)

The frequency of a wire is given by
$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{g}{l}}$
or $f\propto \frac{1}{\sqrt{l}}$
$\therefore \frac{f^{\prime }}{f}=\frac{\sqrt{l}}{{\sqrt{l}}^{\prime }}$
$\Rightarrow \frac{f^{\prime }}{f}=\frac{\sqrt{l}}{\sqrt{2l}}=\frac{1}{\sqrt{2}}$
$\Rightarrow f^{\prime }=0.7f\left[\because l^{\prime }=2l\right]$