Question

Two vibrating strings of the same material but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency ${\nu }_1$ and the other with frequency ${\nu }_2$. The ratio $\frac{{\nu }_1}{{\nu }_2}$ is

• A. 2
• B. 4
• C. 8
• D. 1

Answer 1

The correct option is D 1

Let the density of material be $\rho$ which is same for both the wires.

Therefore, mass per unit lengths will be:

$\Rightarrow {\mu }_1=\frac{\pi {\left(2r\right)}^{2}L\rho }{L}=4\pi r^2\rho$

$\Rightarrow {\mu }_2=\frac{\pi {\left(r\right)}^{2}2L\rho }{2L}=\pi r^2\rho$

Velocity of transverse wave in a stretched string with tension $T$ and mass per unit length $\mu$ is given by:

$\Rightarrow v=\sqrt{\frac{T}{\mu }}$

For the two wires, we have: $T_1=T_2=T$

$\Rightarrow v_1=\sqrt{\frac{T}{{\mu }_1}}=\sqrt{\frac{T}{4\pi r^2\rho }}$

$\Rightarrow v_2=\sqrt{\frac{T}{{\mu }_2}}=\sqrt{\frac{T}{\pi r^2\rho }}$

In the fundamental mode, the wavelength is twice the length of the wire.

${\lambda }_1=2L\Rightarrow {\nu }_1=\frac{v_1}{2L}=\frac{1}{2L}\sqrt{\frac{T}{4\pi r^2\rho }}=\frac{1}{4L}\sqrt{\frac{T}{4\pi r^2\rho }}{\lambda }_2=4L\Rightarrow {\nu }_2=\frac{v_2}{4L}=\frac{1}{4L}\sqrt{\frac{T}{\pi r^2\rho }}=\frac{1}{4L}\sqrt{\frac{T}{4\pi r^2\rho }}\Rightarrow {\nu }_1={\nu }_2\Rightarrow \frac{{\nu }_1}{{\nu }_2}=1$