Answer
$1.39\;kg$
Work Step by Step
Given wave Eq. is
$y=(0.10\;m)(\sin\frac{\pi x}{2})\sin12\pi t\;.................(1)$
From Eq. $1$, we obtain
$k=\frac{\pi}{2}\;rad/m$ and $\omega=12\pi\;rad/s$
Therefore, the speed of the waves on the rope is
$v=\frac{\omega}{k}$
or, $v=\frac{12\pi\times2}{\pi}\;m/s$
or, $v=24\;m/s$
The rope oscillates in a second-harmonic standing wave pattern
Therefore,
$f=2\frac{v}{2L}$
or, $\frac{\omega}{2\pi}=2\frac{\omega}{k\times2\times L}$
or, $L=\frac{2\pi}{k}$
or, $L=\frac{2\pi\times 2}{\pi}\;m$
or, $L=4\;m$
Therefore, the length of the rope is $4\;m$
Let, the mass of the rope is $M\;kg$
Thus, $\mu=\frac{M}{L}=\frac{M}{4}\;kg/m$
The rope is under tension $T=200\;N$
Therefore,
$v=\sqrt {\frac{T}{\mu}}$
or, $24=\sqrt {\frac{4\times200}{M}}$
or, $M=\frac{4\times200}{24^2}$
or, $M=1.39\;kg$
Therefore, the mass of the rope is $1.39\;kg$
