Q.A person standing between two parallel hills fires a gun and hears the first echo after $t_1$ sec and the second echo after $t_2$ sec. The distance between the two hills is

Reflection of Waves at a Boundary

Imagine you're holding one end of a long, loose rope that's tied to a wall at the other end. If you flick your wrist upward, a hump travels down the rope. When that hump hits the wall, something interesting happens: the rope can't move the wall, so the hump comes back — but now it's upside down. That's reflection with a phase reversal.

Now imagine the rope is tied to a light, loose string instead of a wall. The hump reaches the joint, and again it comes back — but this time it stays right-side up. That's reflection without a phase change.

The core idea is simple: a wave reflects when it hits a boundary where the medium changes. What happens to the wave's phase depends entirely on whether the new medium is "harder" or "softer" for the wave to travel through.

The Precise Statement

When a wave travelling in one medium encounters a boundary with another medium:

• Rigid (denser) boundary — the wave reflects with a phase reversal of $\pi$ radians (180°). The reflected wave is inverted relative to the incident wave.
• Free (rarer) boundary — the wave reflects without any phase change. The reflected wave is upright, same as the incident wave.

Why Does This Happen?

Think about what a "rigid" boundary means physically. The wall is fixed — it cannot move. When the rope's displacement tries to push the wall upward, Newton's third law says the wall pushes back with an equal and opposite force. That force generates a reflected pulse that is inverted. The displacement of the rope at the wall must always be zero (the wall doesn't move), so the incident and reflected waves must cancel exactly at that point. The only way that works is if the reflected wave is the mirror image of the incident wave.

For a free boundary, the end is light and can move freely. There's no opposing force. The end simply follows the wave's motion, and the reflected pulse comes back with the same orientation. At the boundary, the displacement is maximum — the incident and reflected waves add constructively.

A Concrete Example: Sound Waves

Sound waves in a pipe work exactly the same way.

A closed end of a pipe is like a rigid boundary — air molecules can't move at the wall, so the reflected sound wave is inverted (phase change of $\pi$). An open end is like a free boundary — air can move freely, so the reflected wave comes back with no phase change.

This is why a closed pipe produces a note one octave lower than an open pipe of the same length: the boundary conditions force different standing wave patterns.

The Mathematical Picture

For a wave $y_i = A \sin(kx - \omega t)$ incident on a boundary at $x = 0$: …