A wire having a linear density of 5 g/m 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 highest frequency at which the same wire resonates is 490 Hz. Find the length of the wire. 

Let the frequency 420 Hz corresponds to p^th harmonic. The formula for p^th harmonic is given by 

${\mathrm{n}}_{\mathrm{p}}=\frac{\mathrm{p}}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{T}}{\mathrm{\mu }}\right)}$

Hence, $420=\frac{\mathrm{p}}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{T}}{\mathrm{\mu }}\right)}$          …(i)

For the next higher frequency p is (p + 1), hence 

$490=\frac{\mathrm{p}+1}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{T}}{\mathrm{\mu }}\right)}$                 …(ii)

From Eqs. (i) and (ii), we have $\frac{490}{420}=\frac{\mathrm{p}+1}{\mathrm{p}}$

Solving we get, p = 6

Substituting the value p in Eq. (i), we get 

$420=\frac{6}{2\mathrm{l}}{\left(\frac{450}{5\times {10}^{-3}}\right)}^{1/2}\Rightarrow \mathrm{l}=2.14\mathrm{m}$