Question

If the equation of a reflected wave from a free surface is y = 2sin(4x+3t). What will be the equation of the original wave?

• A. y = 2cos(4x+3t)
• B. y = 2sin(4x-3t)
• C. y = 2sin(3t-4x)
• D. y = 2sin(4x-3t)

Answer 1

Answer: (B), (D)

y = 2sin(4x-3t)

Answer 2

To determine the equation of the original wave when the reflected wave from a free surface is given, we need to understand the rules of wave reflection from a free boundary.

Reflection from a Free Surface:

When a wave encounters a free boundary (like a free end of a string), the boundary condition is that the transverse displacement is maximum and the transverse force (and thus the slope $\frac{\partial y}{\partial x}$) is zero. Crucially, for reflection from a free surface, there is no phase change upon reflection. The wave simply reverses its direction of propagation.

Let the original (incident) wave be $y_i$.
Let the reflected wave be $y_r$.

The general form of a sinusoidal wave propagating in the positive x-direction is $y = A\sin(kx - \omega t + \phi_0)$.
The general form of a sinusoidal wave propagating in the negative x-direction is $y = A\sin(kx + \omega t + \phi_0)$.

The given reflected wave from a free surface is $y_r = 2\sin(4x+3t)$.
This wave is propagating in the negative x-direction (due to the $+3t$ term).
Since it is a reflected wave, the original wave must have been propagating in the opposite direction, i.e., in the positive x-direction.

According to the rule for reflection from a free surface, there is no phase change. This means the amplitude and the initial phase of the wave remain the same, only the direction of propagation changes.

If the reflected wave is $y_r = A\sin(kx + \omega t)$, then the incident wave must have been $y_i = A\sin(kx - \omega t)$.

Comparing $y_r = 2\sin(4x+3t)$ with $y_r = A\sin(kx + \omega t)$:
We identify the amplitude $A=2$, wave number $k=4$, and angular frequency $\omega=3$.

Therefore, the original wave (incident wave) must be:
$y_i = A\sin(kx - \omega t)$
$y_i = 2\sin(4x - 3t)$