Question

A wire having a linear mass density $5.0\times {10}^{-3}kg/m$ is stretched between two rigid supports with tension of $450N$. The wire reasonates at frequency $400Hz$. The next higher frequency at which the wire resonates is $500Hz$. The length of the wire will be

• A. $1m$
• B. $4m$
• C. $4.5m$
• D. $1.5m$

Answer 1

The correct option is D $1.5m$

It is a wire fixed at both ends and will have all harmonics.
Given that $400Hz,500Hz$ are two consecutive frequencies

$\therefore 400=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
$\Rightarrow 500=\frac{n+1}{2l}\sqrt{\frac{T}{\mu }}$

Dividing,

$\frac{5}{4}=\frac{n+1}{n}$
$5n=4n+4$
$n=4$

Now
$400=\frac{4}{2l}\sqrt{\frac{450}{5\times {10}^{-3}}}$
$\Rightarrow l=\frac{1}{50}\times \sqrt{9\times {10}^4}$
$=\frac{3\times {10}^2}{200}$
$=1.5m$