Question

A wire having a linear mass density $5\times {10}^{-3}\text{kg/m}$ is stretched between two rigid supports with a tension of $450\text{N}$. The wire resonates at a frequency of $420\text{Hz}$. The next higher frequency at which the same wire resonates is $490\text{Hz}$. The length of the wire will be

• A. $4\text{m}$
• B. $2\text{m}$
• C. $8\text{m}$
• D. $3\text{m}$

Answer 1

The correct option is B $2\text{m}$

The resonance frequency of wire,
$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}.........\left(1\right)$
where, $n=$ number of loops
$T=$ Tension, $\mu =$ mass per unit length.
$\Rightarrow \frac{f}{n}=$ constant

For a given string under a tension $\left(T\right)$,
we have, $f_1{n}_2=f_2{n}_1$
From data given in question,
$\frac{420}{490}=\frac{n}{n+1}\Rightarrow n=6$
$\therefore$ substituting the value of $n=6$, in $\left(1\right)$
we have,
$420=\frac{6}{2L}\sqrt{\frac{450}{5\times {10}^{-3}}}$
$\Rightarrow 420=\frac{3}{L}\times 300$

$\Rightarrow L=\frac{900}{420}=2.1\text{m}\approx 2\text{m}$