Question

A sinusoidal wave travels along a taut string of linear mass density $0.1\text{g/cm}$. The particles oscillate along $y-$ direction and the wave moves in the positive $x-$direction. The amplitude and frequency of oscillation are $2\text{mm}$ and $50\text{Hz}$ respectively. The minimum distance between two particles oscillating in the same phase is $4\text{m}$. The tension in the string is

• A. $4000\text{N}$
• B. $400\text{N}$
• C. $45\text{N}$
• D. $250\text{N}$

Answer 1

The correct option is B $400\text{N}$

Given:
$f=50\text{Hz};d=4\text{m}$

$\mu =0.1\text{g/cm}=0.01\text{kg/m}$

The particles oscillating in same phase are separated by distance $\lambda ,2\lambda ,3\lambda ,4\lambda .....etc$

The minimum distance for same phase is $\lambda$
$\because d_{min}=4\text{m}$ (given)

$\Rightarrow \lambda =4\text{m}$

from $v=f\lambda$

$\Rightarrow v=50\times 4\Rightarrow \sqrt{\frac{T}{\mu }}=200$

Thus,

$\sqrt{\frac{T}{0.01}}=200$

$\Rightarrow \frac{T}{0.01}=4\times {10}^4$

$\therefore T=400\text{N}$

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

Why this question? Tip: A particle of medium completes l cycle in a duration of one time period and during that the wave travels a distance $\lambda$.