Q.Two wires of the same material and same cross section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other 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 …