The fundamental frequency of a wire stretched by $2 \mathrm{kgwt}$ is $100 \mathrm{~Hz}$. The weight…
The fundamental frequency of a wire stretched by $2 \mathrm{kgwt}$ is $100 \mathrm{~Hz}$. The weight required to produce its octave is
$12 \mathrm{kgwt}$
8 kgwt
4 kgwt
$16 \mathrm{kgwt}$
Solution
Octave means the new frequency is $2 n$
The velocity of sound is proportional to the square-root of the tension in the wire.
The frequency is directly proportional to the velocity of sound in the wire:
$\begin{aligned} & \therefore n \propto \sqrt{T} \\ & \Rightarrow \frac{n_2}{n_1}=\sqrt{\frac{T_2}{T_1}} \\ & \therefore T_2=T_1\left[\frac{n_2^2}{n_1^2}\right]=4 T_1=8 \mathrm{kgwt}\end{aligned}$