Oscillations · H2 9749 Topic 10

The SHM Test & The Four Graphs Lab

a = −ω²x · x = x₀ sin ωt · v = ±ω√(x₀² − x²)

Try first:press “a = −k x³” and watch the line bend.

Station 1 — the defining-equation tester

is a ∝ −x ?

running: a = −ω²x — mass on an ideal spring: F = −kx exactly, so a = −(k/m)x.

TRACING…

let the oscillator cover one full swing, from +x₀ to −x₀

angular frequency ω5.00 rad s⁻¹

amplitude x₀1.00 m

fitted gradient

0.0 s⁻²

collecting points…

intercept c

0.00

0.0% of a_max

ω from −gradient

—

slider ω = 5.00

worst stray from line

0.0%

of a_max — under 2% counts as straight

T measured

…

between successive upward zero-crossings

2π/ω

1.257 s

only SHM matches this

ⓘ more

Every candidate is integrated numerically (velocity-Verlet, 600 steps s⁻¹) from rest at x₀, and the dots are simply where it went — one dot per frame. The non-SHM laws are scaled so they reach the SAME acceleration at the SAME extreme as true SHM, so all the loci share both end points: only the shape in between can give them away. Drag the amplitude with a non-SHM law selected and watch T change — SHM is the only one of the five whose period does not depend on amplitude.

Station 2 — the four graphs, built by taking gradients

v = dx/dt · a = dv/dt

angular frequency ω5.00 rad s⁻¹

amplitude x₀1.00 m

probe — put the tangent anywheret = 0.339 s = 0.270 T

at the probe: gradient of x–t = -0.63 m s⁻¹ = v, and x = 0.99 m, a = -24.8 m s⁻² (opposite sign to x — always inward).
v and a are the same sign → speeding up, so the mass is moving TOWARDS equilibrium.

at t = 0

x = 0.00 m

v = 5.00 m s⁻¹ — at equilibrium, moving: sin

T peak → peak

1.257 s

2π/ω = 1.257 s

v peak vs x peak

build v–t

measured peak-to-peak once the graph exists

a peak vs x peak

build a–t

half a cycle apart

v_max measured

5.00 m s⁻¹

ωx₀ = 5.00 — at x = 0

a_max measured

25.0 m s⁻²

ω²x₀ = 25.0 — at x = ±x₀

a ÷ x from the graphs

build a–t

live at the cursor

x = 0.00 m

v = 5.00 · a = 0.0

ⓘ more

The v–t and a–t curves here are not typed in from formulas: each point is a central-difference gradient of the curve above it, computed at that instant. That is the exam technique — given one graph, get the next by taking gradients. Every quantity in the panel on the right is measured off those curves by scanning them (peak positions, maxima, a ÷ x), so the phase results are found, not recalled. Flip the sin/cos toggle: all three graphs shift together, the gaps between their peaks do not move, and the a ÷ x ratio is untouched — the start convention sets where the clock starts, nothing else.

Station 3 — the v–x ellipse

(x/x₀)² + (v/ωx₀)² = 1

angular frequency ω5.00 rad s⁻¹

amplitude x₀ (release displacement)1.00 m

damping γ — a = −ω²x − 2γv0.00 s⁻¹

probe displacement x0.50 m = 0.50 x₀

ω√(x₀² − x²) at the probe

4.330 m s⁻¹

what the formula predicts

measured |v| passing that x

waiting for a pass…

the mass has not been there yet

measured |v| at x = 0

…

ωx₀ = 5.000 — the v-axis intercept

measured |x| when v = 0

…

x₀ = 1.000 — the x-axis intercept

shape of the trace

closed ellipse

same loop every cycle

damped ω

5.000 rad s⁻¹

√(ω² − γ²) — equals ω when γ = 0

ⓘ more

The dashed curve is the equation v = ±ω√(x₀² − x²) plotted for every x at once; the solid curve is where the numerically integrated mass has actually been. With γ = 0 they lie on top of one another, and the two intercepts are measured by catching |v| at the instant x crosses 0 and |x| at the instant v crosses 0 — not read off the formula. There is no time axis anywhere on this plot, which is exactly why v = ±ω√(x₀² − x²) is the tool for questions that give you a position and no clock. Turn damping up and the closed loop becomes a spiral about the same centre: x₀ and ωx₀ shrink together, so the prediction at a fixed probe x drifts above the measured value.

What this confirms

a = −ω²x is the DEFINING equation — LO (d). Station 1 never assumes it. Each candidate law is integrated numerically and the (x, a) pairs it visits are plotted as they happen. Only a = −ω²x lays them on a straight line through the origin with negative gradient; the fitted gradient comes back as −ω² and √(−gradient) reproduces the ω on the slider.
The a–x straight line is the TEST, not a decoration. a = −kx³, a = −k|x|x and a = −k·sign(x) all point inward and all oscillate, yet every one of them leaves the line: the cubic bows away from it, the sign law collapses to two horizontal levels with a jump at the origin. Restoring force alone is not SHM.
Both clauses of the definition, separated. Proportionality is the straightness; direction is the negative gradient. The sign(x) candidate keeps the second clause and fails the first, which is exactly the mark lost by writing “a ∝ x” without “always directed towards a fixed point”.
Restoring force F = −mω²x. Multiply the a-axis of the SHM trace by m and it is the F–x graph the notes demand: straight, through the origin, negative gradient, so F is always directed back at equilibrium.
The small-angle approximation, earned. The pendulum candidate is the exact α = −(g/L) sin θ. At θ₀ = 2.5 rad the locus is a visible arc and the verdict is NOT SHM; shrink θ₀ and the arc straightens onto the dashed a = −ω²x line while the printed sin θ₀ vs θ₀ error falls below 0.5%. A pendulum is only approximately simple harmonic, and the slider says how approximately.
Only SHM is isochronous — LO (c). T is measured between successive upward zero-crossings. For a = −ω²x it stays at 2π/ω however far you pull the mass; drag the amplitude with any other candidate selected and T moves — the pendulum slows down at large swings, the cubic speeds up.
x = x₀ sin ωt and x = x₀ cos ωt as solutions — LO (e), and why the choice is forced. The start-convention toggle changes nothing about the motion, only the instant the clock starts: the sin setting opens at x = 0 with v at its largest, the cos setting opens at x = x₀ with v = 0. The t = 0 readout states both, so sin-or-cos is decided by the release condition rather than by memory.
v = v₀ cos ωt built by taking gradients — LO (f), (g). Station 2 constructs v–t from tangents drawn on x–t and a–t from tangents drawn on v–t: central-difference gradients, dropped onto the graph below and joined. The measured peak of the constructed v–t equals ωx₀ and the measured peak of a–t equals ω²x₀ — one factor of ω per differentiation.
Phase relationships found, not recalled — LO (g). The gap between the v peak and the nearest x peak is measured off the graphs as 0.250 T ⇒ φ = π/2 with v leading, and the a peak sits 0.500 T from the x peak ⇒ φ = π, antiphase. The scanned ratio a ÷ x comes back constant at −ω², which is the antiphase drawn as an equation.
Reading directions off an x–t graph. The movable probe reports the tangent gradient as v and the sign of a against the sign of x. When v and a share a sign the mass is speeding up and heading for equilibrium; when they oppose, it is slowing on its way out. That pairing is the whole “when are v and a in opposite directions” question.
v = ±ω√(x₀² − x²) and the v–x ellipse — LO (f). The simulated trail lands on the dashed formula curve, cutting the x-axis at ±x₀ (release point, v = 0) and the v-axis at ±ωx₀ (first pass through equilibrium), traversed clockwise. The probe returns two v values of equal size and opposite sign at one x, which is what the ± means. Both intercepts are verified by catching |v| as x crosses 0 and |x| as v crosses 0.
Damping turns the ellipse into a spiral — LO (i). With γ > 0 the loop no longer closes: same centre, intercepts shrinking together, and the measured |v| at a fixed probe x now falls short of ω√(x₀² − x²) because x₀ itself is decaying. The formula assumes a constant amplitude.