What you can learn with this simulation, topic by topic. Follow each Try it to see it on screen.
1. The medium sets the wave speed
The speed v of a wave on the string depends only on the tension T and the linear density \mu. Changing the amplitude or frequency does not change it. Four times the tension doubles the speed; four times the density halves it.
v=\sqrt{\frac{T}{\mu}}
Try it In pulse mode with a fixed end, turn on the stopwatch. Measure the time t for a pulse to return to the source and compare v=\frac{2L}{t} (L=12\,\mathrm{m}) with the displayed speed. Raising the tension from 1.6\,\mathrm{N} to 6.4\,\mathrm{N} cuts the round trip from 6\,\mathrm{s} to 3\,\mathrm{s}.
2. Speed, frequency and wavelength
During one period \frac{1}{f} of the source, the wave advances one wavelength \lambda. Since the medium fixes the speed, doubling the frequency halves the wavelength.
v=f\lambda
Try it Oscillate mode, open end, damping 0: read the crest spacing on the grid at 1\,\mathrm{Hz} and 2\,\mathrm{Hz} (4\,\mathrm{m}, 2\,\mathrm{m}). Probes one wavelength apart give overlapping graphs; half a wavelength apart, opposite ones.
3. The medium oscillates in place
As a wave passes, each part of the string moves only up and down; it does not travel sideways. What travels is the shape of the disturbance and its energy. The motion is perpendicular to the direction of travel, so this is a transverse wave.
The string on screen is the displacement at one instant, y(x); a probe graph is one point over time, y(t). They look alike but their horizontal axes differ.
Try it In bead view, follow one red bead. As a pulse passes it rises and returns, without moving to the right.
4. Reflection at the end
A fixed end cannot move, so the reflected displacement is inverted (phase change \pi). A free end moves freely, so the reflection comes back upright. The open end lets the wave leave without reflection, as if the string went on forever.
Try it Send the same pulse with a fixed end and then a free end and compare the returning pulse. At reflection the free-end ring rises to about twice the pulse height, because incident and reflected waves overlap.
5. Superposition and standing waves
Two waves of equal amplitude and frequency travelling in opposite directions add up to a standing wave.
y=A\sin(kx-\omega t)+A\sin(kx+\omega t)=2A\sin kx\,\cos\omega t
Each position has a fixed amplitude 2A\lvert\sin kx\rvert and the pattern does not move sideways. Zero amplitude marks a node, maximum an antinode. Neighbouring nodes are \frac{\lambda}{2} apart; a node and the next antinode \frac{\lambda}{4}.
Try it Oscillate mode, fixed end, press Standing-wave setup. Compare probe 1 (antinode) with probe 2 (node) and check the node spacing \frac{\lambda}{2}=2\,\mathrm{m} on the grid.
6. Normal modes: standing waves that need no energy supply
A string with fixed ends allows only standing waves that satisfy both ends: with both ends fixed its length must be a whole number of half wavelengths; with one free end, an odd number of quarter wavelengths (n=1,2,3,\dots).
With zero damping the string's energy is conserved. Released from a normal-mode shape, the string keeps oscillating with no energy supplied, trading kinetic and potential energy: all kinetic when flat, all potential at maximum bend. A plucked guitar string sounds this way. With damping \gamma the amplitude decays as e^{-\gamma t/2}.
Try it Normal-mode mode with a fixed end: change n to 1, 2, 3 and count nodes (n+1 including the ends). Release with damping 0 and watch the amplitude readout stay constant; then add damping and compare.
7. Resonance and damping: a standing wave growing from rest
When the source frequency equals a natural frequency f_n, the source does net work every cycle and the amplitude grows: resonance. Without damping it grows without bound; with damping it stops where supply equals loss. Off resonance the source gets back in each cycle what it gives, so the average work is 0.
With damping, a standing wave forms by itself even from a string at rest: the start-up transient dies out, leaving periodic motion at the source frequency. Without damping the transient never dies out, so a pure standing wave is never reached from rest.
| Damping \gamma\ (\mathrm{s^{-1}}) | Antinode ÷ source amplitude | Time to settle |
| 0 | Grows without bound | Never settles |
| 0.05 | 13.4 | about 120\,\mathrm{s} |
| 0.1 | 6.7 | about 60\,\mathrm{s} |
| 0.3 | 2.3 | about 20\,\mathrm{s} |
| 0.6 | 1.3 | about 11\,\mathrm{s} |
Fixed end, v=4\,\mathrm{m/s}, resonance f=1\,\mathrm{Hz}: time from rest until the string stays within 5\% of the steady state, computed with this simulator.
The settling time, about \frac{6}{\gamma}, is how long the transient takes to decay, on or off resonance. A crisp standing wave needs a resonance frequency and small damping: larger antinodes and deeper nodes, but a longer wait. Tension and density shift the resonances through v but barely change the settling time.
With damping even nodes keep moving a little: the wave weakens on its way to the end and back, so the two opposite waves are not equally strong. For a node at distance d from the reflecting end, its amplitude is about \frac{\gamma d}{2v} of the antinode's; nodes nearer the source are shallower.
Try it Oscillate mode, fixed end, damping 0, 1\,\mathrm{Hz}: the amplitude keeps growing; at 0.9\,\mathrm{Hz} it does not. Standing-wave setup picks the nearest resonance, \gamma=0.2\,\mathrm{s^{-1}}, amplitude 5\,\mathrm{cm} and 2\times speed, and returns the string to rest. After Start, a standing wave with antinodes about 3.4 times the source amplitude (about 17\,\mathrm{cm}) settles in about 30\,\mathrm{s} (about 15\,\mathrm{s} on screen). Probe 2 sits on the deepest node near the reflecting end, probe 1 on the neighbouring antinode.