Question

A wire of linear mass density $9\times {10}^{–3}\text{kg/m}$ is stretched between two rigid supports under a tension of $360\text{N}$. The wire resonates at frequency $210\text{Hz}$. The next higher frequency at which the same wire resonates is $280\text{Hz}$. The number of loops produced in first case will be

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\mu =9\times {10}^{-3}\text{kg}$
$T=360\text{N}$
Frequency $=\frac{nv}{2L}$
Hence $\frac{nv}{2L}=210$
$\frac{(n+1)v}{2L}=280$
$\frac{n}{n+1}=\frac{21}{28}\Rightarrow 28n=21n+21$
$3n=21$
$n=3$
In first case
$\lambda =\frac{2L}{3}$
$L=\frac{3\lambda }{2}$
A loop is formed in length $\frac{\lambda }{2}$
Hence no. of loops are $3$