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 $450\text{Hz}$. 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 $0.7\text{m}$

Let resonance frequencies are $n$ and $(n+1)$.
$n\frac{v}{2l}=400\text{Hz}...\left(1\right)$
$(n+1)\frac{v}{2l}=450\text{Hz}...\left(2\right)$
From Equation (1) and (2)
$\Rightarrow \frac{n}{n+1}=\frac{400}{450}$
$9n=8n+8$
$n=8$
From $\left(1\right)$
$400=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
$400=\frac{8}{2l}\sqrt{\frac{490}{0.1}}$
$400=\frac{4}{l}\left[70\right]$
$l=0.7\text{m}$
Hence option (D) is correct.