Question

A transverse wave is described by $y=\left(0.02m\right)\sin \left[\left(1.0m^{-1}\right)x+\left(30s^{-1}\right)t\right]$ propagates on a stretched string having a linear mass density of $1.2\times {10}^{-4}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$. find the tension in the string.

Answer 1

Step1: Given data and assumptions.

Transverse wave equation, $y=\left(0.02m\right)\sin \left[\left(1.0m^{-1}\right)x+\left(30s^{-1}\right)t\right]$

Linear mass density, $\rho =1.2\times {10}^{-4}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Step2: Find the tension in the string.

Formula used:

$T=\rho v^2$

Where, $T$ is the tension, $\rho$ is the density, and $v$ is the velocity

We have,

$y=\left(0.02m\right)\sin \left[\left(1.0m^{-1}\right)x+\left(30s^{-1}\right)t\right]$ …..$\left(i\right)$

As compared equation $\left(i\right)$ from the standard equation, $y=A\sin \left[kx+\omega t\right]$ we get,

Angular frequency, $\omega =30s^{-1}$

And constant, $k=1.0m^{-1}$

Therefore,

$\omega =30$

$\Rightarrow$$2\pi f=30$ $\left[\because \omega =2\pi f\right]$

$\Rightarrow$ $f=\frac{30}{2\pi }$ …….$\left(ii\right)$

Also,

$\Rightarrow$ $k=1.0$

$\Rightarrow$$\frac{2\pi }{\lambda }=1.0$

$\Rightarrow$ $\lambda =\frac{2\pi }{1.0}$ …….$\left(iii\right)$

Then the velocity of the wave in the stretched string.

$v=\lambda f$

$\Rightarrow$$v=\frac{2\pi }{1}\times \frac{30}{2\pi }$

$\Rightarrow$$v=30\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ ..…..$\left(iv\right)$

Now,

$\Rightarrow$$T=\rho v^2$

$\Rightarrow$$T=1.2\times {10}^{-4}\times {30}^2$

$\Rightarrow$$T=0.108N$

Hence, the tension in the string is $0.108N$