A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal refl

IV. Exercises · Q3

Q. A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2 s. Calculate the depth of the sea in each case and also compute the difference in height between the two cases.

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Concept understanding — Reflection of Waves at a Boundary

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.

Important

The phase change is exactly $\pi$ (half a wavelength) at a rigid boundary, and zero at a free boundary. This is a fixed result — no exceptions for standard mechanical waves.

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.

Note

| Boundary type | What happens at the boundary | Phase change |

|:---|:---|:---:|

| Rigid (fixed end) | Displacement is always zero | $\pi$ (inverted) |

| Free (open end) | Displacement is maximum | 0 (same orientation) |

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.

Watch out

| A common mistake is to think "denser" means "heavier" in the everyday sense. In wave reflection, "denser" means the wave speed is lower in the second medium. For a rope, a heavier rope is denser; for sound, a rigid wall is effectively denser because the wave can't propagate through it.

The Mathematical Picture

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

Rigid boundary: The reflected wave is $y_r = -A \sin(kx + \omega t)$ — the minus sign is the $\pi$ phase shift.
Free boundary: The reflected wave is $y_r = A \sin(kx + \omega t)$ — no minus sign.

The total displacement at the boundary ( $x=0$ ) is:

Rigid: $y(0,t) = A\sin(-\omega t) - A\sin(\omega t) = 0$ (always zero)
Free: $y(0,t) = A\sin(-\omega t) + A\sin(\omega t) = 0$ (wait — that's also zero?)

Actually, for a free boundary, the condition is that the slope (derivative with respect to $x$ ) is zero at the boundary, not the displacement. The displacement is maximum there. The math works out cleanly, but the key takeaway is the phase rule.

y_{\text{reflected}} = \begin{cases} -y_{\text{incident}} & \text{at a rigid boundary} \\ +y_{\text{incident}} & \text{at a free boundary} \end{cases}

The One Thing to Remember

When a wave hits a boundary, it always reflects. The only question is whether the reflected wave flips upside down or not. If the boundary is fixed (rigid, closed, denser), it flips. If the boundary is free (open, rarer), it doesn't. That's the entire concept — everything else follows from this simple rule.

Reflection of waves at a boundary is a foundational NCERT Class 11 Physics topic in the Waves chapter, and 'reflection of waves at rigid and free boundary class 11 physics' or 'phase change on reflection important questions' are common exam-prep searches. This rigid-versus-free boundary distinction also underlies standing-wave and organ-pipe questions that appear regularly in JEE Main and NEET physics.

Step 1. SONAR depth is found from the round-trip time: $d=v\,t/2$ , where t is the time for the signal to travel down and echo back. Using the chapter's own sea-water speed of sound, $v=1533\ \text{m/s}$ (Table 11.2).

Step 2. First case: $t_1=3.5\ \text{s}$ , so $d_1=v\,t_1/2=1533\times3.5/2=1533\times1.75=2682.75\ \text{m}$ .

Step 3. Second case (after the ship moves 100 km, sea floor depth differs): $t_2=2\ \text{s}$ , so $d_2=v\,t_2/2=1533\times2/2=1533\ \text{m}$ .

Step 4. The difference in depth between the two cases is $\Delta d=d_1-d_2=2682.75-1533=1149.75\ \text{m}$ .

✓Final answer

d_1=2682.75\ \text{m}

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