Also number of loops = Number of antinodes.
Hence, with \(5\) antinodes and hanging mass of \(9 kg\).
We have \(p = 5\) and \(T = 9g\) ==> \({n_1} = \frac{5}{{2l}}\sqrt {\frac{{9g}}{m}} \)
With \(3\) antinodes and hanging mass \(M\)
We have \(p = 3\) and \(T = Mg\)
==> \({n_2} = \frac{3}{{2l}}\sqrt {\frac{{Mg}}{m}} \)
\( \because n_1 = n_2\)
==> \(\frac{5}{{2l}}\sqrt {\frac{{9g}}{m}} = \frac{3}{{2l}}\sqrt {\frac{{Mg}}{m}} \)==> \(M = 25\, kg.\)
$ {z_1} = A\sin (kx - \omega \,t) $ , $ {z_2} = A\sin (kx + \omega \,t) $ , $ {z_3} = A\sin (ky - \omega \,t) $ .