Question

Transverse waves are generated in two uniform steel wires A and B of diameters ${10}^{-3}m$ and $0.5\times {10}^{-3}m$ respectively, by attaching their free end to a vibrating source of frequency 500 Hz. The ratio of the wavelengths if they are stretched with the same tension is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $2$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\frac{1}{2}$

The density $\rho$ of a wire of mass M, length L and diameter d is given by
$\rho =\frac{4M}{\pi d^{2}L}=\frac{4\mu }{\pi d^2}$ where $\mu$ is mass per unit length of the wire.
Now, $v_A=\sqrt{\frac{T}{{\mu }_A}}$ and $v_B=\sqrt{\frac{T}{{\mu }_B}}$
$\therefore \frac{v_A}{v_B}=\sqrt{\frac{{\mu }_B}{{\mu }_A}}=\frac{d_B}{d_A}$
But $v_A=n{\lambda }_A$ and $v_B=n{\lambda }_B$, n being the frequency of the source.
Hence, $\frac{{\lambda }_A}{{\lambda }_B}=\frac{v_A}{v_B}=\frac{d_B}{d_A}=\frac{0.5\times {10}^{-3}}{{10}^{-3}}=0.5$, which is choice (a).