Question

Standing waves are set up in a string of length $240cm$ clamped horizontally at both ends. The separation between any two consecutive points where displacement amplitude is $3\sqrt{2}cm$ is $20cm$. The standing waves were set by two travelling waves of equal amplitude of $3cm$. The overtone in which the string is vibrating will be

Answer 1

$2A\sin kx=3\sqrt{2}$
$2\times 3\sin kx=3\sqrt{2}$
$\sin kx=\frac{1}{\sqrt{2}}$
$\frac{2\pi }{\lambda }x=\frac{\pi }{4};\frac{3\pi }{4}$
$x_n=\frac{\lambda }{8}{x}_{n+1}=\frac{3\lambda }{8}$
Distance between consecutive points
$=\frac{3\lambda }{8}-\frac{\lambda }{8}=\frac{\lambda }{4}$
$\frac{\lambda }{4}=20cm$
$\lambda =80cm$
So,
$(n+1)\frac{\lambda }{2}=240$
$(n+1)\frac{80}{2}=240$
or $n+1=6$
$n=5$
So, string is vibrating in fifth overtone.