⊣ Wave Reflection
A pulse and a wave at a boundary: fixed end, free end and transition into another medium, with reflection and transmission coefficients.
Boundary:
Wave:
Animation speed×1,0
Examples (partial transmission):
Coefficients at the boundary
—
Reflection r (amplitude)
—
Transmission t (amplitude)
—
Reflected energy R = r²
—
Transmitted energy 1 − r²
—
Phase shift on reflection
—
Standing wave ratio (1 + |r|) / (1 − |r|)
Fixed boundary: r = −1, the displacement at the boundary is always zero Free boundary: r = +1, the displacement at the boundary doubles
Two media: r = (Z₁ − Z₂) / (Z₁ + Z₂) = (1 − k) / (1 + k), t = 2Z₁ / (Z₁ + Z₂) = 2 / (1 + k), k = Z₂ / Z₁
For a string: Z = √(T μ) = μ v, t = 1 + r, r² + (Z₂ / Z₁) t² = 1
Two media: r = (Z₁ − Z₂) / (Z₁ + Z₂) = (1 − k) / (1 + k), t = 2Z₁ / (Z₁ + Z₂) = 2 / (1 + k), k = Z₂ / Z₁
For a string: Z = √(T μ) = μ v, t = 1 + r, r² + (Z₂ / Z₁) t² = 1
- The light blue line is the sum of the incident and reflected waves to the left of the boundary. Turn on “Show incident and reflected waves separately” to see them as dashed lines. At a fixed boundary the reflected wave is inverted, at a free one it is not.
- In “Partial transmission” mode the second medium is shaded on the right. The purple wave is the part that has passed through. It can have a larger amplitude than the incident wave (t up to 2) but carries less energy: r² + (Z₂/Z₁)·t² = 1.
- Relative impedance k = Z₂ / Z₁. For k > 1 the second medium is “heavier” (r < 0, the pulse is reflected inverted), for k < 1 it is “lighter” (r > 0). As k → ∞ the boundary behaves as fixed, as k → 0 as free, and at k = 1 there is no reflection.
- The “Continuous sine wave” view shows the standing wave produced by the superposition of the incident and reflected waves. At |r| = 1 there are stationary nodes; at smaller values they do not reach zero, and this is described by the standing wave ratio.
Two strings with different linear densities, joined and stretched by the same force. Enter the tension and the densities to get the wave speeds, the relative impedance and the reflection and transmission coefficients.
Tension T, N
Density of the first string μ₁, g/m
Density of the second string μ₂, g/m
Examples:
—
Speed v₁, m/s
—
Speed v₂, m/s
—
Relative impedance k = Z₂ / Z₁
—
Reflection r
—
Transmission t
—
Reflected / transmitted energy, %
v = √(T / μ), Z = √(T μ), k = Z₂ / Z₁ = √(μ₂ / μ₁)
r = (1 − k) / (1 + k), t = 2 / (1 + k)
r = (1 − k) / (1 + k), t = 2 / (1 + k)
Coefficients as a function of the relative impedance k = Z₂ / Z₁
| k | Reflection r | Transmission t | Reflected energy | Transmitted energy |
|---|---|---|---|---|
| 0,1 | 0,82 | 1,82 | 67 % | 33 % |
| 0,3 | 0,54 | 1,54 | 29 % | 71 % |
| 0,5 | 0,33 | 1,33 | 11 % | 89 % |
| 1 | 0,00 | 1,00 | 0 % | 100 % |
| 1,6 | -0,23 | 0,77 | 5 % | 95 % |
| 2 | -0,33 | 0,67 | 11 % | 89 % |
| 3 | -0,50 | 0,50 | 25 % | 75 % |
| 10 | -0,82 | 0,18 | 67 % | 33 % |
| ∞ (fixed) | −1,00 | 0,00 | 100 % | 0 % |
| 0 (free) | +1,00 | — | 100 % | 0 % |
Wave Reflection Online: Fixed and Free Boundary, Partial Transmission
A free interactive tool that shows how a wave pulse is reflected from a boundary. At a fixed end the pulse is inverted, at a free end it is reflected without inversion, and at a transition into another medium part of the wave is reflected and part continues onward. Change the relative impedance and the speed in the second medium, switch between a pulse and a continuous sine wave, and watch the reflection and transmission coefficients, the energy and the standing wave ratio. The calculator computes these quantities for two strings of different density.
Key concepts
Reflection coefficient
The ratio of the amplitude of the reflected wave to that of the incident wave. A negative value means inversion (a 180° phase shift).
r = (1 − k) / (1 + k)
Medium impedance
A characteristic of the wave resistance of a medium. For a string Z = √(Tμ). Reflection occurs where the impedance changes; with equal impedances the wave passes through without reflection.
Z₁ = Z₂ → r = 0
Standing wave on reflection
The superposition of the incident and reflected sine waves. With total reflection stationary nodes form; with partial reflection the nodes are incomplete.
SWR = (1 + |r|) / (1 − |r|)
Where it is used
- Transmission lines and coaxial cables: matching the load so the signal is not reflected
- Ultrasound diagnostics: reflection at boundaries of tissues with different acoustic impedance
- Anti-reflective coatings on optics and the reflection of light at the interface of two media
- Seismology: reflection and transmission of waves at the boundaries of Earth’s layers
- Musical instruments and room acoustics
- Laboratory work on waves on a string and a spring
How to use
- Choose the fixed boundary and watch the pulse invert on reflection
- Switch to the free boundary: the pulse is reflected with the same sign, and the end moves with a doubled swing
- Turn on “Partial transmission” and move the relative impedance: the reflection, transmission and energy change
- Find k = 1, where the reflection disappears
- Switch “Wave” to the sine wave and look at the standing wave
- In the “Calculator” tab, enter the tension and densities of your own strings
Frequently asked questions
Why is the pulse inverted at a fixed end?
At a fixed end the string cannot be displaced: y = 0 at every moment. This is possible only if the reflected wave at the boundary point exactly cancels the incident one, that is, has the opposite sign. So the reflected pulse is inverted, r = −1.
What happens at a free end?
A free end can move, but the tension force there is zero, which means the slope of the string is zero. The reflected wave has the same sign, r = +1, and at the boundary itself the incident and reflected waves add up, so the displacement of the end is twice the amplitude of the pulse.
What is relative impedance and how does it determine reflection?
For a string the impedance is Z = √(Tμ). For two strings under the same tension the relative impedance is k = Z₂/Z₁ = √(μ₂/μ₁). The reflection coefficient is r = (1 − k)/(1 + k), and the transmission coefficient is t = 2/(1 + k). If k = 1, there is no reflection. If k ≫ 1, the boundary is almost fixed, and if k ≪ 1, almost free.
Can the transmitted wave have a larger amplitude than the incident one?
Yes, when the second medium is “lighter” (k < 1), the transmission coefficient t = 2/(1 + k) is greater than 1, up to 2. This does not violate energy conservation: the transmitted wave carries an energy flux (Z₂/Z₁)·t², and together with the reflected r² it always equals 1.
Why is r negative for k > 1 in the simulation, when the original version has it differently?
In this tool k = Z₂/Z₁ and r = (1 − k)/(1 + k). Then as k → ∞ the boundary behaves as fixed with r → −1, and as k → 0 as free with r → +1. In the original demonstration the formula r = (k − 1)/(k + 1) gives the opposite sign, so it has been corrected here.