The equation of a wave on a string oflinear mass density $0.04$ $kgm^{-1}$ is given by
$y = 0.02sin\left[ {2\pi \left( {\frac{t}{{0.04\left( s \right)}} - \frac{x}{{0.50\left( m \right)}}} \right)} \right]m$ The tension in the string is .... $N$
AIEEE 2010, Diffcult
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$y=0.02(m) \sin \left[2 \pi\left(\frac{t}{0.04(s)}\right)-\frac{x}{0.50(m)}\right]$
But $y=a \sin (\omega t-k x)$
$\therefore \omega=\frac{2 \pi}{0.04} \Rightarrow v=\frac{1}{0.04}=25 H z$
$k=\frac{2 \pi}{0.50} \Rightarrow \lambda=0.5 \mathrm{m}$
$\therefore$ velocity, $\mathrm{v}=v \lambda=25 \times 0.5 \mathrm{m} / \mathrm{s}=12.5 \mathrm{m} / \mathrm{s}$
Velocity on a string is given by
$v=\sqrt{\frac{T}{\mu}} \therefore T=\mathrm{v}^{2} \times \mu=(12.5)^{2} \times 0.04=6.25 \mathrm{N}$
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