Interactive physics simulator
Reflection of Waves
Explore how wave pulses interact with boundaries. Study fixed-end inversions, free-end amplitude overshoots, and transmission and reflection dynamics across string density boundaries.
Wave Reflection Laboratory
Trigger a single wave pulse or generate continuous waves. Observe incoming, reflected, and superimposed waves.
Live Wave Reflection Telemetry
- Wave State
- Incident + Reflected
- Boundary Type
- Fixed End (Clamped)
- Phase Change
- 180° Inversion (π rad)
- Wave Speed
- v = 4.00 m/s
- Boundary Coefficients
- R = -1.00, T = 0.00
Introduction to Wave Reflection
In physics, the reflection of waves describes the behavior of a wave when it encounters a boundary or an obstacle and bounces back into the original medium. This phenomenon occurs in all wave types—including transverse waves on a string, sound waves in air, and light waves transitioning between optical media.
The manner in which a wave reflects depends fundamentally on the boundary conditions it encounters. There are two primary types of boundaries: Fixed Boundaries and Free Boundaries.
Fixed End vs. Free End Reflection
The mechanical forces acting at the boundary dictate the phase of the reflected wave:
- Fixed-End Reflection (Inverted): If the end of a string is clamped securely to a wall, the string cannot move at that point. When an upward wave pulse arrives, it pulls upward on the clamp. By **Newton\'s third law**, the clamp exerts an equal and opposite downward reaction force on the string. This reaction force generates a reflected pulse that travels back **upside down (inverted)**. This flip is equivalent to a **180° (or π rad) phase change**.
- Free-End Reflection (Upright): If the end of a string is attached to a ring that slides frictionlessly up and down a vertical rod, it can move freely. The rod exerts no downward reaction force on the string. When the upward wave pulse reaches the rod, the free end behaves as an antinode, overshooting to **twice the wave\'s initial amplitude**. The pulse then reflects back **upright (non-inverted)**, representing **0° phase change (0 rad)**.
Wave Transmission & Boundary Impedance
When a wave reaches a junction between two different media (such as a thin string spliced to a thick rope), it encounters a change in the medium\'s **impedance** (its resistance to wave propagation). In this case, the wave is partially reflected and partially transmitted:
- Transitioning from Light to Heavy String (Thin to Thick): The heavier string has high mass density, presenting higher inertia. The boundary behaves similarly to a **fixed end**. Part of the wave energy is reflected **inverted**, while part of the energy is transmitted **upright** into the heavy string, propagating slower with a shorter wavelength.
- Transitioning from Heavy to Light String (Thick to Thin): The lighter string has lower inertia, presenting less resistance. The boundary behaves similarly to a **free end**. Part of the wave energy is reflected **upright**, and part is transmitted **upright** into the light string, propagating faster with a longer wavelength.
The Mathematical Boundary Equations
The fractions of wave amplitude reflected and transmitted are quantified by the Reflection Coefficient (R) and Transmission Coefficient (T):
T = 2v2 / (v2 + v1)
Where:
- v1: The wave speed in the initial medium.
- v2: The wave speed in the second medium.
- R: Reflection coefficient. A negative R value signifies an inverted reflected wave.
- T: Transmission coefficient. T is always positive, meaning the transmitted wave is always upright.
Note that the coefficients satisfy the boundary continuity relation: R + 1 = T.
Solved Examples
A wave pulse with an amplitude of +15 cm travels down a string and reaches a fixed support boundary. Describe: (a) what happens to the pulse at the boundary, (b) the amplitude of the reflected pulse, and (c) the phase shift experienced.
View Step-by-Step Solution
- Part (a): At a fixed support boundary, the string cannot move. The incoming pulse exerts an upward force on the support. By Newton's third law, the support exerts an equal and opposite downward reaction force on the string. This reflects the pulse upside down (inverted).
- Part (b): Assuming an ideal elastic medium with no energy loss, the magnitude of the amplitude remains identical, but its orientation is inverted. The reflected amplitude is y = -15 cm.
- Part (c): An inversion represents a complete upside-down flip of the wave profile, which corresponds to a phase shift of π radians (or 180°).
**Final Answer:** Inverted, amplitude = -15 cm, phase shift = 180° (π rad)
A transverse wave propagates along a light string with a velocity of 8.0 m/s. It hits a boundary junction where it is tied to a heavier string where the wave velocity drops to 4.0 m/s. Calculate: (a) the reflection coefficient R, and (b) the transmission coefficient T at the boundary interface.
View Step-by-Step Solution
- Identify variables: incident speed v1 = 8.0 m/s, and transmitted speed v2 = 4.0 m/s.
- Part (a): Use the reflection coefficient formula: R = (v2 - v1) / (v2 + v1).
- Substitute values: R = (4.0 - 8.0) / (4.0 + 8.0) = -4.0 / 12.0 = -1/3 ≈ -0.33.
- The negative sign indicates that the reflected wave is inverted (phase shift of 180°).
- Part (b): Use the transmission coefficient formula: T = 2v2 / (v2 + v1).
- Substitute values: T = 2 · 4.0 / (4.0 + 8.0) = 8.0 / 12.0 = 2/3 ≈ +0.67.
- The positive sign indicates that the transmitted wave propagates upright (no phase shift).
**Final Answer:** R ≈ -0.33 (inverted), T ≈ +0.67 (upright)
A wave pulse propagates from a heavy rope (wave speed v1 = 3.0 m/s) into a lighter rope (wave speed v2 = 6.0 m/s). If the incident pulse amplitude is +10 cm, calculate: (a) the reflection coefficient R, (b) the amplitude of the reflected pulse, and (c) the amplitude of the transmitted pulse.
View Step-by-Step Solution
- Identify variables: incident speed v1 = 3.0 m/s, transmitted speed v2 = 6.0 m/s, and incident amplitude Ai = +10 cm.
- Part (a): Solve for R: R = (v2 - v1) / (v2 + v1) = (6.0 - 3.0) / (6.0 + 3.0) = 3.0 / 9.0 = +1/3 ≈ +0.33.
- The positive sign means the reflected pulse is upright (acts like a free boundary).
- Part (b): The reflected amplitude is Ar = R · Ai = (1/3) · 10 cm = +3.33 cm.
- Part (c): Solve for T: T = 2v2 / (v2 + v1) = 2 · 6.0 / (6.0 + 3.0) = 12.0 / 9.0 = +4/3 ≈ +1.33.
- The transmitted amplitude is At = T · Ai = (4/3) · 10 cm = +13.33 cm (peaking due to the boundary transition).
**Final Answer:** R ≈ +0.33, Areflected ≈ +3.33 cm, Atransmitted ≈ +13.33 cm
Common Misconceptions & Pitfalls
- Misconception: The frequency of a wave changes when it transmits into a new medium.
**Reality:** No. The frequency of a wave is determined solely by the original source generator. When transitioning to a new medium, wave speed and wavelength change proportionally, but frequency remains strictly constant. - Misconception: Energy is lost when the transmission coefficient T is greater than 1.
**Reality:** No. When a wave goes from heavy to light string, T can be larger than 1 (meaning the transmitted amplitude exceeds the incident amplitude). This is because the light string is much easier to displace. However, because the mass density of the light string is low, the total power/energy transmitted remains strictly conserved. - Misconception: Sound wave reflections only occur off solid walls.
**Reality:** Sound waves reflect off any boundary where the acoustic impedance changes. This includes junctions between warm and cold air masses, boundaries between different gas compositions, or interfaces between water and air.
Practice Questions
Question 1
Why is the amplitude of a transmitted wave pulse at a boundary between a heavy and light rope sometimes larger than the incident pulse amplitude (T > 1)? Does this violate conservation of energy?
Show Explanation
No, it does not violate energy conservation. The transmission coefficient T = 2v₂ / (v₁ + v₂) can be greater than 1 when waves move into a lighter medium (where v₂ > v₁). This happens because the lighter string has less mass per unit length (lower inertia), meaning it is much easier to displace, allowing the wave to peak higher. However, the energy of a wave depends on both amplitude and the medium properties (Energy ∝ μ · A² · v). Since the light medium has a much smaller mass density μ, the total energy carried by the transmitted pulse remains strictly less than or equal to the incident pulse energy.
Question 2
Describe what happens to the wavelength and frequency of a wave pulse when it reflects off a boundary, and when it transmits across a boundary into a different medium.
Show Explanation
For reflection: The wave remains in the original medium, so its wave speed v remains constant. The frequency f is locked to the source, so it remains constant. Since λ = v / f, the wavelength of the reflected wave is identical to the incident wave. For transmission: The wave enters a new medium, changing its speed. The frequency remains constant. As a result, the wavelength changes proportionally (λ ∝ v): if the wave enters a slower medium, its wavelength shortens; if it enters a faster medium, its wavelength stretches.
Question 3
Contrast fixed-end and free-end reflections in terms of the phase and displacement at the boundary point.
Show Explanation
At a fixed-end boundary: The displacement at the boundary is locked at zero (a node). The incoming and reflected waves interfere destructively at the boundary point, and the wave pulse reflects inverted (180° phase shift). At a free-end boundary: The boundary point is free to move (an antinode). The incoming and reflected waves interfere constructively at the boundary point, causing the free end to displace to twice the incoming pulse amplitude, and the pulse reflects upright (0° phase shift).
Frequently Asked Questions
- What is wave reflection?
- Wave reflection occurs when a wave travels through a medium, encounters a boundary or obstacle, and bounces back into the original medium.
- Why does a wave invert at a fixed boundary?
- At a fixed end, the wave exerts an upward force on the clamp. The clamp exerts an equal and opposite downward force on the wave (Newton's Third Law). This reaction force flips the wave pulse, creating a 180° (π rad) phase shift.
- Why does a wave stay upright at a free boundary?
- At a free end (like a loop sliding on a vertical rod), the end can move freely. There is no downward restoring reaction force from the boundary, so the pulse reflects upright with no phase shift (0 rad).
- What happens to wavelength during wave reflection?
- The wavelength remains unchanged. Because the reflected wave travels in the same medium, its speed remains constant, and its frequency is unchanged, meaning Wavelength = Speed / Frequency is constant.
- What happens when a wave transitions across a medium boundary?
- Part of the wave is reflected back into the first medium, and part is transmitted into the second medium. The frequency remains constant on both sides, but the speeds and wavelengths differ.
- What are the reflection (R) and transmission (T) formulas?
- R = (v₂ - v₁) / (v₂ + v₁) and T = 2v₂ / (v₂ + v₁), where v₁ is the speed in the first medium and v₂ is the speed in the second medium.