Question

A wire having a linear density $0.1\text{kg/m}$ is kept under a tension of $490\text{N}$. It is observed that it resonates at a frequency of $400\text{Hz}$ and the next higher frequency of $450\text{Hz}$. Find the length of the wire.

• A. $0.4\text{m}$
• B. $0.6\text{m}$
• C. $0.49\text{m}$
• D. $0.7\text{m}$

Answer 1

The correct option is D $0.7\text{m}$

For a wire fixed at two ends, resonance frequencies are given by $f_n=\frac{nv}{2l}$
Given:
$n\frac{v}{2l}=400\text{Hz}$ ...(1)
& $(n+1)\frac{v}{2l}=450\text{Hz}$ ...(2)

From equation (1) and (2),
$\frac{n}{n+1}=\frac{400}{450}$
We know that, $v=\sqrt{\frac{T}{\mu }}$. Here, $T$ and $\mu$ (linear mass density) are constant.

$\Rightarrow 9n=8n+8$
or $n=8$

From equation (1),
$8\times \frac{1}{2l}\sqrt{\frac{490}{0.1}}=400$
$\Rightarrow l=\frac{8}{2}\times \frac{70}{400}=0.7\text{m}$