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 pth harmonic. The formula for pth harmonic is given by np=p2lTμHence, 420=p2lTμ          …(i)For the next higher frequency p is (p + 1), hence 490=p+12lTμ                 …(ii)From Eqs. (i) and (ii), we have 490420=p+1pSolving we get, p = 6Substituting the value p in Eq. (i), we get 420=62l4505×10−31/2⇒l=2.14m