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A transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$ is represented by the equation $y=0.2 \sin (15 x+60 t)$
where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is
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where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is
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Verified Answer
The correct answer is:
0.48
Equation of wave, $y=0.2 \sin (1.5 x+60 t)$ Comparing with standard equation,
$\begin{aligned}
& y=A \sin (k x+\omega t) \\
& k=1.5, \omega=60 \mathrm{rad} / \mathrm{s}
\end{aligned}$
$\therefore$ Velocity of wave, $v=\frac{\omega}{k}=\frac{60}{1.5}=40 \mathrm{~m} / \mathrm{s}$
Velocity of wave in a stretched string,
$v=\sqrt{\frac{T}{m}}$
where $m=$ linear density
$T=\text { tension in the string }$
So,
$\begin{aligned}
T & =v^2 m \\
& =(40)^2 \times 3 \times 10^{-4}=0.48
\end{aligned}$
$\begin{aligned}
& y=A \sin (k x+\omega t) \\
& k=1.5, \omega=60 \mathrm{rad} / \mathrm{s}
\end{aligned}$
$\therefore$ Velocity of wave, $v=\frac{\omega}{k}=\frac{60}{1.5}=40 \mathrm{~m} / \mathrm{s}$
Velocity of wave in a stretched string,
$v=\sqrt{\frac{T}{m}}$
where $m=$ linear density
$T=\text { tension in the string }$
So,
$\begin{aligned}
T & =v^2 m \\
& =(40)^2 \times 3 \times 10^{-4}=0.48
\end{aligned}$
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