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Sandbox Physics

Mass–Spring Oscillator

What sets the period of a mass bouncing on a spring? Mass and spring stiffness set the period: four times the mass doubles it. Amplitude does not change it. Hanging the spring vertically only moves the equilibrium point down; gravity leaves the stiffness, and with it the period, unchanged.

Drag a spring mass and compare energy exchange, the phase-space ellipse and rotating phasor. Time noisy camera crossings and try vertical suspension.

Interactive modelMass–Spring Oscillator
Kinetic energy28.8 mJ28.8\,\mathrm{mJ}
Potential about equilibrium28.8 mJ28.8\,\mathrm{mJ}
Camera crossing period1.57 s1.57\,\mathrm{s}
Ideal model period1.57 s1.57\,\mathrm{s}

Four pictures of the same oscillation

BackgroundAt the turning point a linear spring stores the energy. Near equilibrium the carriage carries it as motion. Plotting displacement with momentum, or with a rotated velocity coordinate, reveals the same repeating exchange.

Why it mattersOpenStax University Physics Volume 1, Section 15.1 derives simple harmonic motion. This experiment connects its exact solution to independent noisy camera crossings.

Focus question
Which changes alter the period, and which only move the starting point on the orbit?
One-sentence intuition
Mass and stiffness set the period. Amplitude sets the size of the orbit; release phase chooses where the motion starts.

Core mathematical model

Motion relative to equilibrium

mq¨+kq=0,ω0=k/m,T0=2π/ω0m\ddot q+kq=0,\quad\omega_0=\sqrt{k/m},\quad T_0=2\pi/\omega_0

Position is positive rightward on the track and downward when hanging. The equilibrium-location control translates the entire support.

Release amplitude and phase

q=Acos⁡(ω0t+φ),v=−Aω0sin⁡(ω0t+φ)q=A\cos(\omega_0t+\varphi),\qquad v=-A\omega_0\sin(\omega_0t+\varphi)

A quarter-turn phase starts at equilibrium with a kick. Neither phase nor amplitude changes the ideal period.

The ellipse and the circular phasor

E=p22m+kq22,p=mv,q2+(−v/ω0)2=A2E=\frac{p^2}{2m}+\frac{kq^2}{2},\quad p=mv,\quad q^2+(-v/\omega_0)^2=A^2

Momentum against displacement makes an ellipse. The phasor uses negative velocity divided by frequency and equal length scales, so it turns counterclockwise.

The camera clock

t↑,j=a+jT^+rjt_{\uparrow,j}=a+j\widehat T+r_j

Interpolate rising crossings in the camera samples after a resolved negative excursion. Fit crossing time against index; the slope is the period, with a conditional regression standard error.

Hanging does not change the linear restoring law

Δℓstatic=mg/k,ΔUs=12kq2+mgq,ΔUg=−mgq\Delta\ell_{\rm static}=mg/k,\quad\Delta U_s=\tfrac12kq^2+mgq,\quad\Delta U_g=-mgq

Gravity shifts equilibrium. Its energy change cancels the elastic linear term, leaving the potential plotted above.

Common difficulties

Force and velocity do different jobs

Typical misconceptionThe spring force is largest where the carriage is fastest.

Better mental modelAt a turning point speed is zero and restoring force is largest. At equilibrium the spring’s net restoring force is zero while speed is largest.

Hanging equilibrium is stretched

Typical misconceptionGravity must make the clock slower.

Better mental modelIt stretches the spring to a new equilibrium. For this ideal linear law the stiffness about that point stays unchanged.

Run the experiment

  1. 01

    Follow one quarter turn

    Pause, then move Replay time through a quarter period.

    What to observe: Elastic potential becomes kinetic energy while the phase-space point moves along the ellipse.
  2. 02

    Change the clock

    Compare A heavier carriage with A quicker spring.

    What to observe: More mass lengthens the period; more stiffness shortens it. The camera and model periods track the same change.
  3. 03

    Rotate the apparatus

    Choose Hang it vertically and adjust gravity in the details.

    What to observe: Static extension changes, but the period at fixed mass and stiffness stays the same.
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