Equation of travelling wave on a stretched string of linear density $5\,g/m$ is $y = 0.03\,sin\,(450\,t -9x)$ where distance and time are measured in $SI$ united. The tension in the string is ... $N$
JEE MAIN 2019, Medium
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$y=0.03 \sin \left[450\left(t-\frac{9 x}{450}\right)\right]$
So, $\mathrm{v}=\frac{450}{9}=50 \mathrm{m} / \mathrm{s}$
Also, $v=\sqrt{\frac{T}{\lambda}}$
$\Rightarrow 50=\sqrt{\frac{T}{5 \times 10^{-3}}}$
$\Rightarrow T=2500 \times 5 \times 10^{-3}=12.5 \mathrm{N}$
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