Question

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

• A. $10\text{N}$
• B. $7.5\text{N}$
• C. $12.5\text{N}$
• D. $5\text{N}$

Answer 1

The correct option is C $12.5\text{N}$

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

Comparing it with standard equation of wave, we get
$\omega =450,k=9$

$\therefore v=\frac{\omega }{k}=\frac{450}{9}=50{\text{ms}}^{-1}$

The velocity of the travelling wave on a stretched string is given by,

$v=\sqrt{\frac{T}{\mu }}\Rightarrow \frac{T}{\mu }=2500$

$\text{linear mass density},\mu =5{\text{g m}}^{-1}=5\times {10}^{-3}{\text{kg m}}^{-1}$

$\Rightarrow T=2500\times 5\times {10}^{-3}$

$\Rightarrow T=12.5\text{N}$

Hence, option $\left(C\right)$ is correct.