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Reflection of waves looks like a one-line rule: phase change of π at a fixed end, none at a free end. Questions then ask you to write the reflected wave's equation, sketch a pulse after reflection, or find what happens where a light string meets a heavy one, and the rule alone is not enough. This post explains where the rule comes from and works three problems.
What does the syllabus say?
NCERT Class 11 Chapter 14, Waves, section 14.6, states that "a travelling wave or pulse suffers a phase change of π on reflection at a rigid boundary and no phase change on reflection at an open boundary". It gives the open end of an organ pipe as an example of a non-rigid boundary. The CBSE 2026-27 curriculum and the JEE (Main) 2026 syllabus both list "reflection of waves" just before standing waves. Check your exam year's syllabus.
Why does a fixed end flip the wave?
NCERT gives two explanations. First, a rigid end cannot move, so the incident and reflected displacements must add to zero there at every instant. That is only possible if the reflected wave is inverted. Second, the arriving pulse pulls on the wall, and by Newton's third law the wall pulls back the opposite way, launching an inverted pulse.
At a free end, such as a string tied to a light ring sliding on a smooth rod, nothing holds the end down. The end overshoots, reaching twice the pulse's height, and the reflected pulse comes back upright.
Worked example 1: writing the reflected wave
A wave y = a sin(ωt − kx) travels in +x towards a fixed end at x = 0. Write the reflected wave.
The reflected wave travels in −x, so it has the form a sin(ωt + kx + δ).
At x = 0 the total must be zero for all t: a sin ωt + a sin(ωt + δ) = 0, so δ = π.
Reflected wave: y = a sin(ωt + kx + π) = −a sin(ωt + kx).
Add them: y = a sin(ωt − kx) − a sin(ωt + kx) = −2a cos(ωt) sin(kx). At x = 0, sin 0 = 0, so the fixed end is a node, as it must be.
For a free end at x = 0, δ = 0 and the sum is 2a sin(ωt) cos(kx), which puts an antinode at the end. These two results are the starting point for the harmonics in our guide to standing waves in strings and pipes.
Worked example 2: a light string joined to a heavy one
Most boundaries are neither fully fixed nor fully free. Take two strings under the same tension joined at a knot, with wave speeds v1 and v2. Matching the displacement and the slope at the knot (a step beyond the NCERT text) gives:
reflected amplitude / incident amplitude = (v2 − v1)/(v1 + v2)
transmitted amplitude / incident amplitude = 2v2/(v1 + v2)
Let the second string have 4 times the mass per unit length. Since v = √(T/μ), v2 = v1/2.
Reflected ratio = (v1/2 − v1)/(3v1/2) = −1/3. The minus sign means the reflected pulse is inverted, as at a fixed end, but only a third as tall.
Transmitted ratio = v1/(3v1/2) = 2/3, and upright.
Check: 1 + (−1/3) = 2/3, so the displacements match at the knot.
Worked example 3: heavy string to light string
Reverse it: the wave starts on the heavy string and the second string has a quarter of the mass per unit length, so v2 = 2v1.
Reflected ratio = (2v1 − v1)/(3v1) = +1/3: upright, like a free end.
Transmitted ratio = 4v1/(3v1) = 4/3.
The transmitted pulse is taller than the incident one. That does not break energy conservation, because the light string carries less energy for the same amplitude. Rule of thumb: going into a slower (denser) medium inverts the reflection; going into a faster one does not.
Why do students mix these up?
Wittmann, Steinberg and Redish (1999) studied student reasoning about mechanical waves and found that individual students use both correct and incorrect mental models to make sense of the same physical situation. Reflection invites this, because "fixed", "rigid", "closed" and "denser" all point the same way and are easy to blur. Tie each case back to what the boundary point can do.
Which mistakes cost marks?
Flipping the direction but not the sign. The reflected wave needs both: +kx and the phase change.
Assuming every reflection inverts. Free ends and fast-to-slow junctions do not.
Forgetting that frequency stays the same across a junction. Speed and wavelength change; frequency does not.
A practice routine
Sketch a single upright pulse before and after reflection at a fixed end and a free end.
Rewrite example 1 for an incident wave a sin(kx − ωt) and check the node.
Work the junction ratios for μ2 = 9μ1 and μ2 = μ1/9.
Revisit the travelling wave equation if signs of kx and ωt still confuse you.
To watch pulses flip at a boundary, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Next step: write the reflected wave for a free end at x = 0 and confirm the antinode yourself.
