Question

A transverse wave described by

$y=\left(0·02\mathrm{m}\right)\sin \left(1·0{\mathrm{m}}^{-1}\right)x+\left(30{\mathrm{s}}^{-1}\right)t$

propagates on a stretched string having a linear mass density of $1·2\times {10}^{-4}\mathrm{kg}{\mathrm{m}}^{-1}$. Find

Answer 1

Given,
Wave equation, $y=\left(0·02\mathrm{m}\right)\sin \left(1·0{\mathrm{m}}^{-1}\right)x+\left(30{\mathrm{s}}^{-1}\right)t$

$$\begin{gathered} \mathrm{Let}: \\ \mathrm{Mass}\mathrm{per}\mathrm{unit}\mathrm{length},m=1.2\times {10}^{-4}\mathrm{kg}/\mathrm{m} \\ \mathrm{From}\mathrm{the}\mathrm{wave}\mathrm{equation},\mathrm{we}\mathrm{have}: \\ k=1{\mathrm{m}}^{-1}=\frac{2\pi }{\lambda } \\ \mathrm{And}, \\ \omega =30{\mathrm{s}}^{-1}=2\pi f \end{gathered}$$

Velocity of the wave in the stretched string is given by

$$\begin{gathered} \nu =\lambda f=\frac{\omega }{k}=\frac{30}{1} \\ \Rightarrow v=30\mathrm{m}/\mathrm{s} \\ \mathrm{We}\mathrm{know}: \\ v=\sqrt{\frac{T}{m}} \\ \Rightarrow 30=\sqrt{\left(\frac{T}{1.2\times {10}^{-4}}\right)} \\ \Rightarrow T=108\times {10}^{-3}=0.108\mathrm{N} \end{gathered}$$

So, the tension in the string is 0.108 N.