Question

The wave function of a triangular wave pulse is defined by the relation below at time $t=0\text{sec}$
$y=\{\begin{array}{cc}mx& \text{for}0\le x\le \frac{a}{2}\\ -m(x-a)& \text{for}\frac{a}{2}\le x\le a\\ 0& \text{every where else}\end{array}$

The wave pulse is moving in the + X direction in a string having tension $T$ and mass per unit lengh $\mu$. The total kinetic energy present with the wave pulse is

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

Answer 1

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

Kinetic energy present with the wave pulse:
${\int }_0^{\frac{a}{2}}\frac{1}{2}\mu dx{\left(\frac{\delta y}{\delta t}\right)}^2+{\int }_{\frac{a}{2}}^a\frac{1}{2}\mu dx{\left(\frac{\delta y}{\delta t}\right)}^2$
$\left(\frac{\delta y}{\delta t}\right)=\frac{mdx}{dt}$ for $0\le x\le \frac{a}{2}$
$\left(\frac{\delta y}{\delta t}\right)=-\frac{mdx}{dt}$ for $\frac{a}{2}\le x\le a$
where $\frac{dx}{dt}=\sqrt{\frac{T}{\mu }}$
$KE={\int }_0^{\frac{a}{2}}\frac{1}{2}{m}^{2}Tdx+{\int }_{\frac{a}{2}}^a\frac{1}{2}{m}^{2}Tdx$
$KE=\frac{m^{2}T}{2}(\frac{a}{2}-0)+\frac{m^{2}T}{2}(a-\frac{a}{2})$
$KE=\frac{m^{2}Ta}{2}$