Let tension be \(T\)
\(f_1=\sqrt{\frac{T}{\mu}} \times \frac{1}{2 l}=256\)
\(f_2=\sqrt{\frac{T+10}{\mu}} \times \frac{1}{2 l}=320\)
\(\sqrt{\frac{T}{T+10}}=\frac{256}{320}\)
\(\frac{T}{T+10}=\frac{(16)^2}{(16)^2 \times(20)^2}\)
or \(\frac{T}{T+10}=\frac{16^2}{(20)^2}\)
or \(400 T=256 T^2+2560\)
or \(144 T=2560\)
or \(T=\frac{2560}{144}\)
or \(T=\frac{2560}{16 \times 9}\)
or \(T=\frac{160}{9}\) Newton
\(=\frac{16}{9} \,kg -w t\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$
$y_1=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), \quad y_2=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
પરિણામી તરંગનું સમીકરણ ક્યું છે.