Q.A wire has linear density 4.0 × 10⁻³ kg/m. It is stretched between two rigid supports with a tension of 360 N. The wire resonates at a frequency of 420 Hz and 490 Hz in two successive modes. Find the length of the wire.

A string of length l, linear density (mass per unit length) m, held under tension T between two rigid supports, has a NODE at each fixed end (the same boundary condition, at both ends, as an open pipe's antinode/antinode pair, just with the roles of node and antinode swapped). The fundamental mode -- plucking exactly at the centre -- fits half a wavelength along the string, $\lambda=2l$, giving fundamental frequency $n=\dfrac{v}{2l}=\dfrac{1}{2l}\sqrt{T/m}$ (using the transverse wave speed on a stretched string, $v=\sqrt{T/m}$). Because both ends share the SAME boundary condition (node/node), a string -- exactly like an open pipe -- supports ALL harmonics as overtones: $n_p=pn$ for $p=1,2,3,\ldots$, obtained physically by damping the string at successive equally-spaced internal points and plucking between them to force additional internal nodes (two loops for the second harmonic/first overtone, three loops for the third harmonic/second over …