Question

The wavefunction of a triangular wave pulse is defined by the relation below at time $t=0\text{sec}$
$$y=\{\begin{array}{cc}mx& \text{for}0\le x\le \frac{a}{2}\\ -m(x-a)& \text{for}\frac{a}{2}\le x\le a\\ 0& \text{everywhere else}\end{array}$$
The wave pulse is moving in the $+X$ direction in a string having tension $T$ and mass per unit length $\mu$. The total kinetic energy present in the wave pulse is.

• A. $\frac{m^{2}Ta}{2}$
• B. $\frac{3m^{2}Ta}{2}$
• C. $m^{2}Ta$
• D. None of the above

Answer 1

The correct option is A $\frac{m^{2}Ta}{2}$

Kinetic energy present with the wave pulse:

$K=\underset{0}{\overset{\frac{a}{2}}{\int }}dx(\frac{1}{2}\times \mu \times (\frac{\partial y}{\partial t}{)}^2)+\underset{\frac{a}{2}}{\overset{a}{\int }}dx(\frac{1}{2}\times \mu \times (\frac{\partial y}{\partial t}{)}^2)$

From the data given in the question,

$\frac{\partial y}{\partial t}=m\frac{dx}{dt};\text{for}0\le x\le \frac{a}{2}$

$\frac{\partial y}{\partial t}=-m\frac{dx}{dt};\text{for}\frac{a}{2}\le x\le a$

Also, $\frac{dx}{dt}=\sqrt{\left(\frac{T}{\mu }\right)}$

Thus, $K=\frac{m^{2}Ta}{4}+(\frac{m^{2}Ta}{2}-\frac{m^{2}Ta}{4})$

$\Rightarrow K=\frac{m^{2}Ta}{2}$

Hence, option (a) is the correct answer.