Question

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

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

Answer 1

The correct option is B $12.5N$

Given, travelling wave equation,

$y=0.03\sin (450t-9x)$

Velocity of the wave, $v=\frac{\omega }{k}$

From the wave equation, $\omega =450,k=9$

$\Rightarrow v=\frac{450}{9}=50m/s$

Given, linear density of the string

$\mu =5g/m=5\times {10}^{-3}kg/m$

As we know, speed of the wave $v=\sqrt{\frac{T}{\mu }}$
So, tension in the string $T=v^2\mu$
$T=\left(50{)}^2\right(5\times {10}^{-3})$
$T=12.5N$
Final answer: (𝑏)