Question

A string of length $40\text{cm}$ and linear mass density $0.80\text{g/cm}$ is fixed at both ends and kept under a tension of $32\text{N}$. A wave pulse is produced near one of the ends at $t=0$, as shown in the figure, which makes periodic motion between the ends. What is the time after which the string will have the same shape shown in the figure?

• A. After $0.03\text{sec}$
• B. After $0.06\text{sec}$
• C. After $0.05\text{sec}$
• D. After $0.04\text{sec}$

Answer 1

The correct option is D After $0.04\text{sec}$

Given that,
Length of the string $\left(l\right)=40\text{cm}$
Linear mass density of the string $\left(\mu \right)=0.80\text{g/cm}\text{or}0.080\text{kg/m}$
Tension in the string $\left(T\right)=32\text{N}$

The wave will be inverted (phase reversal) after striking with fixed end.


The image shows that the string will have same shape after $2$ consecutive reflections.

Total distance travelled by wave pulse $x=40+40=80\text{cm}$

Wave speed in the stretched string, $v=\sqrt{\frac{T}{\mu }}$
From the given data,

$v=\sqrt{\frac{32}{0.08}}=20\text{m/s}$

Therefore, time taken to regain initial shape $t=\frac{0.8\text{m}}{20\text{m/s}}=0.04\text{sec}$.

Hence, option (d) is the correct answer.