Question

Equation of travelling wave on a stretched string of linear density $5g/m$ is $y=0.03sin\left(450t9x\right)$ where distance and time are measured is SI units. The tension in the string is :

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

Answer 1

Answer: (B)

$12.5N$

$y=0.03\sin (450t-9x)$
$v=\frac{\omega }{k}=\frac{450}{9}=50m/s$
$v\sqrt{\frac{T}{\mu }}\Rightarrow \frac{T}{\mu }=2500$
$\Rightarrow T=2500\times 5\times {10}^{-3}$
$=12.5N$