Question

A transverse wave propagating on a stretched string of linear density $3\times {10}^{-4}kg{m}^{-4}$ is represented by the equation, $y=\left(0.02\right)\sin \left[1.5x+60t\right]$, where, $x$ is in meter and $t$ is in second. The tension in the string (in newton) is

• A. $0.48$
• B. $0.24$
• C. $1.20$
• D. $1.80$

Answer 1

The correct option is A

$0.48$

Step1: Given data and assumptions.

Transverse equation, $y=\left(0.02\right)\sin \left[1.5x+60t\right]$

Linear mass density, $\rho =3\times {10}^{-4}kg{m}^{-4}$

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,

$y=\left(0.02\right)\sin \left[1.5x+60t\right]$ …..$\left(i\right)$

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

Angular frequency, $\omega =60s^{-1}$

And constant, $k=1.5m^{-1}$

Therefore,

$\omega =60$

$\Rightarrow$$2\pi f=60$ $\left[\because \omega =2\pi f\right]$

$\Rightarrow$ $f=\frac{60}{2\pi }$ …….$\left(ii\right)$

Also,

$\Rightarrow$ $k=1.5$

$\Rightarrow$$\frac{2\pi }{\lambda }=1.5$

$\Rightarrow$ $\lambda =\frac{2\pi }{1.5}$ …….$\left(iii\right)$

Then the velocity of the wave in the stretched string.

$v=\lambda f$

$\Rightarrow$$v=\frac{2\pi }{1.5}\times \frac{60}{2\pi }$

$\Rightarrow$$v=40\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ ..…..$\left(iv\right)$

Now,

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

$\Rightarrow$$T=3\times {10}^{-4}\times {40}^2$

$\Rightarrow$$T=0.48N$

Hence, option A is correct. The tension in the string is $0.48N$