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)

Frequency of vibration in the wire,
$v=\frac{1}{2l}\sqrt{\frac{T}{\mu }}...(i)$
where, $l=$ length of the wire
$T=$ tension in the wire
$\mu =$ mass per unit length
$v^{\prime }=\frac{1}{2l^{\prime }}\sqrt{\frac{T}{{\mu }^{\prime }}}..(ii)$
From Eqs. (i) and (ii), we get
$\frac{v}{v^{\prime }}=\frac{l^{\prime }}{l}\times \sqrt{\frac{{\mu }^{\prime }}{\mu }}=\frac{2l}{l}\times \sqrt{\frac{\rho \pi r^{\prime }2}{\rho \pi r^2}}$
$=\frac{2l}{l}\times \sqrt{{\left(\frac{r^{\prime }}{r}\right)}^2}=2\times \frac{r^{\prime }}{r}=2\times 2=4$
$\Rightarrow v=n=4v^{\prime }\Rightarrow v^{\prime }=\frac{v}{4}$