Question

The equation of a wave is: y = 2sin(4x - 3t) What will be the equation of the reflected wave from a free surface?

Answer 1

y = -2sin(4x+3t)

Answer 2

For a wave on a string, the boundary condition at a free end is that the slope (∂y/∂x) vanishes at the free surface. With the incident wave

$$y_i = 2\sin(4x-3t)$$

assume the reflected wave has the form

$$y_r = 2\sin(4x+3t+\phi).$$

At a free end (say at $x=0$), the net transverse slope must be zero:

$$\left.\frac{\partial}{\partial x}(y_i+y_r)\right|_{x=0} = 0.$$

Differentiate:

$$\frac{\partial y_i}{\partial x} = 8\cos(4x-3t), \quad \frac{\partial y_r}{\partial x} = 8\cos(4x+3t+\phi).$$

At $x=0$:

$$8\cos(-3t) + 8\cos(3t+\phi) = 8\cos3t + 8\cos(3t+\phi) = 0.$$

Dividing by 8 and using the cosine sum identity:

$$\cos 3t + \cos(3t+\phi) = 2\cos\Big(3t+\frac{\phi}{2}\Big)\cos\frac{\phi}{2}=0.$$

For this to hold for all $t$, we require

$$\cos\frac{\phi}{2}=0 \quad\Longrightarrow\quad \frac{\phi}{2}=\frac{\pi}{2}+n\pi \quad \Longrightarrow \quad \phi=\pi+2n\pi.$$

Choosing the principal value $\phi=\pi$, we get

$$y_r = 2\sin(4x+3t+\pi) = -2\sin(4x+3t).$$

Thus, the reflected wave from a free surface is:

$$y = -2\sin(4x+3t).$$

Brief Explanation:

1. Assume reflected wave: $y_r= 2\sin(4x+3t+\phi)$.
2. Apply free-end condition $\frac{\partial}{\partial x}(y_i+y_r)=0$ at $x=0$ to find $\phi=\pi$.
3. Hence, $y_r= -2\sin(4x+3t)$.