Predict how mass changes the period

Count a complete repeat

For the default 1 kg mass, the block starts at +0.200 m at rest, reaches the negative turning point after 1 s, and returns to its original state after 2 s. The period T is 2 s. The frequency f is 1/T = 0.5 Hz, meaning half a complete cycle per second.

Crossing equilibrium twice does not mean completing two cycles. The crossings have opposite velocity. To measure a period, compare events with the same position and phase, such as successive positive turning points or successive rightward equilibrium crossings.

Change inertia while keeping the spring fixed

The restoring acceleration is a = −kx/m. A larger mass has smaller acceleration magnitude at the same displacement for a fixed spring. For an ideal undamped linear spring, the full result is:

T = 2π√(m/k) and f = (1/2π)√(k/m).

Here m is mass in kilograms, k is stiffness in newtons per metre, T is in seconds and f is in hertz. The spring in this lab has k = π² N/m. Changing mass prepares a separate release with the same amplitude; it does not attach extra mass to an already moving block.

Set m = 0.25 kg. The period is 1 s, frequency 1 Hz, and the 4 s plot contains four full cycles. Return to 1 kg: period 2 s, frequency 0.5 Hz, and the same plot contains two cycles. The mass increased by a factor of four, but period increased by √4 = 2, not four. Fixed time axes keep that comparison visible.

At m = 2 kg, period is approximately 2.828427 s. The trace stops at 4 s because that is the selected observation window, not because the oscillator stops. Read the timing ledger when a whole cycle does not line up with integer seconds or the quarter-second control steps.

Distinguish two frequencies

Angular frequency omega = √(k/m) = 2π/T measures phase change in radians per second. Ordinary frequency f counts cycles per second. One cycle contains 2π radians, so omega = 2πf.

For m = 1 kg, omega = π rad/s ≈ 3.141593 rad/s, while f = 0.5 Hz. Calling omega “3.14 Hz” would predict too many cycles. For m = 0.25 kg, omega = 2π rad/s and f = 1 Hz. Compare both numbers with the actual repeats in the graph.

Timing and energy are different comparisons

The stored release energy E = kA²/2 is approximately 0.197392 J for every mass here, because stiffness and amplitude stay fixed. At equilibrium it is all kinetic energy. Since K = mv²/2, a smaller mass reaches a larger speed with that same energy.

The peak speed is A omega. At 1 kg it is about 0.628319 m/s; at 0.25 kg it is about 1.256637 m/s. Thus the lighter block moves twice as fast at equilibrium and completes the cycle twice as often. Energy conservation and the period formula tell a consistent story.

The ideal linear model’s period is independent of amplitude. A larger release amplitude would increase maximum speed and energy without changing its period, as long as Hooke’s law remains valid. Amplitude is fixed in this lab; this statement is a model prediction, not a separate experiment supplied by the controls. Nonlinear springs and damping require additional models. The OpenStax harmonic-motion reference discusses these ideal timing relations.

Transfer to a different oscillator

A 0.5 kg block on a 20 N/m spring has period about 0.993459 s. Predict the period if mass becomes 2 kg, then if stiffness also becomes 80 N/m. State the frequency in each new case.

Check the period transfer

Increasing mass fourfold doubles period to about 1.986918 s, giving frequency about 0.503292 Hz. Increasing stiffness fourfold as well restores the original m/k ratio and period about 0.993459 s, with frequency about 1.006584 Hz. The ratio m/k determines timing; neither parameter alone is sufficient.

For a physical experiment, time several complete cycles and divide by the number of cycles, record the start/end event definition, and estimate timing uncertainty. This calculated trace supplies exact model times; it does not establish measurement skill or quantify a real spring’s departures from the ideal assumptions.