The length and diameter of a metal wire used in sonometer is doubled. The fundamental frequency will change…
The length and diameter of a metal wire used in sonometer is doubled. The fundamental frequency will change from ' $n$ ' to
$\frac{\mathrm{n}}{4}$
n
2n
$\frac{\mathrm{n}}{2}$
Solution
The fundamental frequency is given by
$\mathrm{n}=\frac{1}{2 \mathrm{Cr}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
$\begin{aligned} & \therefore \mathrm{n} \propto \frac{1}{\ell_{\mathrm{r}}} \\ & \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\ell_1 \mathrm{r}_1}{\ell_2 \mathrm{r}_2}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\end{aligned}$
$\therefore \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}$