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(in m)

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 

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