Waves & Sound · Physics

Reflection & Transmission of a Pulse: 3D Simulation

Q: Does a transmitted pulse ever come out inverted?

No. τ = 2v₂/(v₁ + v₂) is always positive, so the transmitted pulse is always upright. Only the reflected pulse can be inverted, and only when the second string is slower (heavier).

A pulse runs along a 3D string of beads and hits a fixed end, a free ring, or a joint with a heavier or lighter string. Watch it come back inverted or upright, see part of it carry on, and compare the reflected and transmitted heights with r = (v₂ − v₁)/(v₁ + v₂) and τ = 2v₂/(v₁ + v₂).

Interactive virtual lab · no download · works on phones and laptops

Chapter: Waves
For: Class 11 · JEE · NEET
Time: ~15 min

Reflection & Transmission of a Pulse is a Pro simulation

Pro opens all 152 virtual labs — 693 simulations with guides and quizzes.

How to use it

1Pick a module from the list (3); set your level at the top right for more controls.
2Pick the pulse shape and size; the pulse sets off by itself. Turn on the ghosts to see the parts.
3Use Predict before you look, then Derive, Questions, Practise and Compare.

What to notice

From a fixed end the pulse returns inverted; from a free end, upright.
Into a slower (heavier) string: reflected inverted, transmitted upright and shorter-wide.
1 + r = τ: the joint moves with both strings.
Energy is shared: r² reflected, (v₁/v₂)τ² transmitted.

For teachers

A ready-made teaching kit comes with this lab

Lesson plan
22 min, 4 steps
Ask the class
3 questions with answers
Quick quiz
6 questions, printable
Key formulas
4 formulas

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About the Reflection & Transmission of a Pulse simulation

What happens when a pulse on a string reaches the end? This 3D reflection and transmission of a pulse simulation sends a hump, triangle or square pulse along a string of beads into three different ends. At a fixed end — a clamp on a wall — it comes back upside down, because the wall pulls the string the other way. At a free end — a light ring sliding on a smooth rod — it comes back the right way up, and the ring overshoots to twice the pulse's height.

The third module joins two strings under the same tension. Going into a heavier, slower string, part of the pulse reflects inverted and part carries on upright but squeezed narrower; going into a lighter, faster string, the reflection is upright and the transmitted pulse is stretched wider and taller. Turn on the ghosts to see the incident and reflected pulses separately before they add up, and compare the heights with r = (v₂ − v₁)/(v₁ + v₂) and τ = 2v₂/(v₁ + v₂) — the method of images and the boundary conditions that JEE builds questions on.

Also covers: reflection of a pulse, fixed end reflection, free end reflection, phase change on reflection, pulse inverted, transmission at a boundary, light string to heavy string, reflection coefficient, transmission coefficient, impedance of a string, method of images, class 11 waves, jee waves.

Frequently asked questions

Why is a pulse inverted when it reflects from a fixed end?

The fixed end cannot move. As the pulse arrives it pulls up on the support, and by Newton's third law the support pulls down on the string, sending back an inverted pulse. Mathematically the boundary condition y = 0 at the end is met by an inverted image pulse travelling the other way.

What happens to a pulse going from a light string into a heavy string?

The wave is slower in the heavy string. Part of the pulse reflects inverted (r = (v₂ − v₁)/(v₁ + v₂) < 0), like a partly fixed end, and part is transmitted upright with amplitude τ = 2v₂/(v₁ + v₂) and a narrower width.

Does a transmitted pulse ever come out inverted?

No. τ = 2v₂/(v₁ + v₂) is always positive, so the transmitted pulse is always upright. Only the reflected pulse can be inverted, and only when the second string is slower (heavier).

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