Question

The length and diameter of a metal wire is doubled. The fundamental frequency of vibration will change from '$n$' to (Tension being kept constant and material of both the wires is same)

• A. $\frac{n}{4}$
• B. $\frac{n}{8}$
• C. $\frac{n}{12}$
• D. $\frac{n}{16}$

Answer 1

Answer: (A)

$\frac{n}{4}$

Fundamental frequency of vibration $n=\frac{v}{2L}\sqrt{\frac{T}{\mu }}$

where $\mu$ is the mass per unit length of the wire i.e. $\mu =\frac{M}{L}$

Mass of the wire $M=\rho \left(\frac{4\pi }{3}{R}^3\right)$

$⟹$ $n=\frac{v}{2L}.\sqrt{\frac{TL}{\rho \frac{4\pi }{3}{R}^3}}$

$⟹$ $n\propto \frac{1}{R\sqrt{LR}}$ .....(1)

Given : $L_2=2L$ $R_2=2R$

From equation (1), we get $\frac{n_2}{n}=\frac{R\sqrt{RL}}{R_2\sqrt{L_2{R}_2}}$

Or $\frac{n_2}{n}=\frac{R\sqrt{RL}}{\left(2R\right)\sqrt{\left(2L\right)\left(2R\right)}}=\frac{1}{4}$

$⟹$ $n_2=\frac{n}{4}$