Question

A wire having a linear density of 0.05 g/cm is stretched between two rigid supports with a tension of 450 N. It is observed that the wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 490 Hz. The length of the wire is $2142\times {10}^{-x}m$. Find x.

Answer 1

Frequency of oscillation of string in $n^{th}$ harmonic is = $\frac{n}{2l}\sqrt{\frac{T}{\mu }}$.

Thus $\frac{490}{420}=\frac{n_2}{n_1}=\frac{n_1+1}{n_1}$

where $n_1,n_2$ are integers.

Thus $n_1=6$

Thus $420=\frac{6}{2l}\sqrt{\frac{450}{0.05\times \frac{{10}^{-3}}{{10}^{-2}}}}$

$⟹l=2.142m$