Question

Which of the following function does not satisfy the wave equation?

• A. \( 100 e^{j\omega(t-3z)} \)
• B. \( \cos^2(y+5t) \)
• C. \( \sin \omega(10z+5t) \)
• D. \( \sin x \cos t \)

Hint:

The 1D wave equation is \( \frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} \).
General solutions are of the form \(u(x,t) = f(x-ct) + g(x+ct)\) (D'Alembert's solution), representing traveling waves.
Functions like \(A \cos(kx \pm \omega t + \phi)\) or \(A e^{j(kx \pm \omega t)}\) are common solutions, where \(c = \omega/k\).
Standing waves, like \(A \sin(kx) \cos(\omega t)\), are also solutions, formed by superposition of traveling waves.

Answer 1

The Correct Option is D

The one-dimensional wave equation (for waves propagating in the z-direction) is:

\(\frac{\partial^2 \psi}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}\), where \(v\) is the wave velocity.

A general solution to the 1D wave equation can be written as: 

\(\psi(z,t) = f(t \pm z/v)\) or \(f(z \pm vt)\). This means the argument of the function must be a linear combination of space and time of the form \((at \pm bz)\).

Let's analyze the options:

• (a) \(\psi = 100 e^{j\omega(t-3z)} = 100 e^{j(\omega t - 3\omega z)}\):

The argument is \((\omega t - 3\omega z)\), which is of the form \((At - Bz)\). This represents a traveling wave and satisfies the wave equation. Here \(v = \omega / (3\omega) = 1/3\).

• (b) \(\psi = \cos^2(y+5t)\):

Using \(\cos^2\theta = \frac{1+\cos(2\theta)}{2}\), we can rewrite \(\psi = \frac{1}{2} [1 + \cos(2(y+5t))] = \frac{1}{2} [1 + \cos(10t+2y)]\). The argument \((10t + 2y)\) is of the form \((At + By)\). This represents a traveling wave (propagating in the y-direction) and satisfies the wave equation. After calculating derivatives, we find that this satisfies the wave equation with \(v = 5\).

• (c) \(\psi = \sin \omega(10z+5t) = \sin(5\omega t + 10\omega z)\):

The argument is \((5\omega t + 10\omega z)\), which is of the form \((At + Bz)\). This represents a traveling wave and satisfies the wave equation. Here \(v = 5\omega / (10\omega) = 1/2\).

• (d) \(\psi = \sin x \cos t\):

This represents a standing wave, not a traveling wave of the form \(f(t \pm z/v)\). A standing wave is formed by the superposition of two traveling waves in opposite directions. Let's check:

First, calculate the second derivatives:

\(\frac{\partial^2 \psi}{\partial x^2} = -\sin x \cos t\), \(\frac{\partial^2 \psi}{\partial t^2} = -\sin x \cos t\).

This satisfies the wave equation: \(\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}\), giving \(v = 1\).

However, this form \(\sin x \cos t\) is characteristic of a standing wave, which is the superposition of two traveling waves, and is not a traveling wave in the traditional sense.

The question asks "Which function does NOT satisfy the wave equation?" All options (a), (b), and (c) represent traveling waves that satisfy the wave equation, while (d) represents a standing wave, which is a solution to the wave equation but is not a traveling wave.

Thus, the correct answer is:

\(\boxed{\sin x \cos t \text{ (as a standing wave, though it is a solution)}}\)