Question

The wave function of a triangular wave pulse is defined by the relation below at time t = 0 sec.

$$y = \begin{cases} mx & \text{for } 0 \leq x \leq \frac{a}{2}\\ -m(x-a) & \text{for } \frac{a}{2} \leq x \leq a\\ 0 & \text{every where else, where m<<1} \end{cases}$$

The wave pulse is moving in the +X direction in a string having tension T and mass per unit length μ. The total energy present with the wave pulse is :-

• A. $\frac{m^2Ta}{2}$
• B. $m^2Ta$
• C. $mTa$
• D. $\frac{mTa}{2}$

Answer 1

Answer: (B)

$m^2Ta$

Answer 2

The wave function of the triangular wave pulse at time $t=0$ is given by:

$$y(x, 0) = \begin{cases} mx & \text{for } 0 \leq x \leq \frac{a}{2}\\ -m(x-a) & \text{for } \frac{a}{2} \leq x \leq a\\ 0 & \text{every where else} \end{cases}$$

The wave pulse is moving in the +X direction. For a travelling wave $y(x, t) = f(x - vt)$, the total energy is the sum of the kinetic and potential energies. The kinetic energy density is $\frac{1}{2} \mu (\frac{\partial y}{\partial t})^2$ and the potential energy density is $\frac{1}{2} T (\frac{\partial y}{\partial x})^2$. For a travelling wave, the kinetic energy density is equal to the potential energy density at every point. This can be shown as follows: $\frac{\partial y}{\partial t} = -v f'(x - vt)$ and $\frac{\partial y}{\partial x} = f'(x - vt)$. So $(\frac{\partial y}{\partial t})^2 = v^2 (f'(x - vt))^2 = v^2 (\frac{\partial y}{\partial x})^2$. Since the wave speed $v = \sqrt{T/\mu}$, we have $v^2 = T/\mu$. Thus, $\frac{1}{2} \mu (\frac{\partial y}{\partial t})^2 = \frac{1}{2} \mu \frac{T}{\mu} (\frac{\partial y}{\partial x})^2 = \frac{1}{2} T (\frac{\partial y}{\partial x})^2$. The total energy density is the sum of the kinetic and potential energy densities, which is $2 \times \frac{1}{2} T (\frac{\partial y}{\partial x})^2 = T (\frac{\partial y}{\partial x})^2$. The total energy of the wave pulse is the integral of the energy density over the region where the wave exists: $E = \int T (\frac{\partial y}{\partial x})^2 dx$. We need to calculate $\frac{\partial y}{\partial x}$ at $t=0$. This is the slope of the wave function $y(x, 0)$. For $0 \leq x < \frac{a}{2}$, $\frac{\partial y}{\partial x} = \frac{d}{dx}(mx) = m$. For $\frac{a}{2} < x \leq a$, $\frac{\partial y}{\partial x} = \frac{d}{dx}(-m(x-a)) = -m$. The slope is not defined at $x=a/2$, but this is a single point and does not affect the integral. So, $(\frac{\partial y}{\partial x})^2 = m^2$ for $0 \leq x \leq a$, except at $x=a/2$. The total energy is the integral over the region $0 \leq x \leq a$: $E = \int_0^a T (\frac{\partial y}{\partial x})^2 dx = \int_0^a T m^2 dx$. Since $T$ and $m^2$ are constants with respect to $x$ in the region $0 \leq x \leq a$, we have: $E = T m^2 \int_0^a dx = T m^2 [x]_0^a = T m^2 (a - 0) = T m^2 a$.

The total energy present with the wave pulse is $m^2Ta$.