Question

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

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

Answer 1

Answer: (B)

12.5 N

Answer 2

$y =0.03 \sin\left(450 t - 9x\right)$
$v= \frac{\omega}{k} =\frac{450}{9} =50 m/s$
$v= \sqrt{\frac{T}{\mu}} \Rightarrow \frac{T}{\mu} =2500$
$\Rightarrow T = 2500 \times5 \times10^{-3}$
$= 12.5 \,N$