A transverse wave propagating on the string can be described by the equation $y=2 \sin (10 x+300 t)$. where $x$ and $y$ are in metres and $t$ in second. If the vibrating string has linear density of $0.6 \times 10^{-3} \,g / cm$, then the tension in the string is .............. $N$
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(b)
$y=2 \sin (10 x+300 t)$
$y=A \sin (k x+\omega t)$
$k=10, \quad \omega=300$
$\mu=0.6 \times 10^{-3}\,g / cm$
$\mu=\frac{0.6 \times 10^{-3} \times 10^{-3}}{10^{-2}}\,kg / m$
$\mu=6 \times 10^{-5}\,kg / m$
$K=\frac{\omega}{V}$
$V=\frac{\omega}{K}=\frac{30 p}{T p}=30\,m / s$
$V=\sqrt{\frac{T}{M}}$
$T=V^2 \times M$
$T=(30)^2 \times 6 \times 10^{-5}$
$=900 \times 6 \times 10^5$
$=5400 \times 10^{-5}$
$=0.0541\,N$
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