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,
$v=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\,\,\,\,...(i)$
where, $l=$ length of the wire
$T =$ tension in the wire
$\mu=$ mass per unit length
$v'=\frac{1}{2 l'} \sqrt{\frac{T}{\mu'}} \,\,\,\,\,..(ii)$
From Eqs. (i) and (ii), we get
$\frac{v}{v'}=\frac{l'}{l} \times \sqrt{\frac{\mu'}{\mu}}=\frac{2 l}{l} \times \sqrt{\frac{\rho \pi r' 2}{\rho \pi r^{2}}}$
$=\frac{2 l}{l} \times \sqrt{\left(\frac{r'}{r}\right)^{2}}=2 \times \frac{r'}{r}=2 \times 2=4 $
$\Rightarrow \,\,\, v =n=4 v' \Rightarrow v'=\frac{v}{4} $