Question

Equation of travelling wave on a stretched string of linear density $5\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$ is $y=0.03\sin \left(450t-9x\right)$ where distance and time are measured in $SI$ units. The tension in the string is:

• A. $12.5N$
• B. $7.5N$
• C. $10N$
• D. $5N$

Answer 1

The correct option is A

$12.5N$

Step 1. Given data

Equation of travelling wave on a stretched string is $y=0.03\sin \left(450t-9x\right)$

Linear density of stretched string is $\mu =5\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Step 2. Finding propagation velocity.

We know that, the propagation velocity is given by $V=\frac{\omega }{\kappa }$

Where, $V$ is the propagation velocity of the wave

$\omega$ is the angular velocity of the wave

$\kappa$ is the wavenumber if the wave equation

Now, compare the given wave equation with $y=A\sin \left(\omega t-\kappa x\right)$ to obtain the values of $\omega$ and $\kappa$

Therefore, $V=\frac{450}{9}=50\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 3. Finding the tension in the string.

We know that tension in the stretched string is given by $T=V^2\times \mu$

Where, $V$ is the propagation velocity of the wave

$\mu$ is the linear density of the string

By substituting the values we get,

$$\begin{gathered} T=2500\times 5\times {10}^{-3} \\ =12.5N \end{gathered}$$

Hence, the correct option is (A).