A motion sensor records acceleration against displacement for an oscillator: (−0.080 m, +5.12 m s⁻²), (−0.040 m, +2.56 m s⁻²), (0, 0), (0.040 m, −2.56 m s⁻²), and (0.080 m, −5.12 m s⁻²). Determine whether the data support SHM, find angular frequency and period, and calculate maximum speed and total energy for mass 0.250 kg. Evaluate how a constant acceleration-sensor zero offset would appear on the graph.

Question

Accepted Answer

The linear negative gradient is −64.0 s⁻², so ω = 8.00 rad s⁻¹ and T = 0.785 s. With A = 0.080 m, v_max = 0.640 m s⁻¹ and E = 0.0512 J. SHM requires a = −ω²x: a straight acceleration–displacement graph through the origin with negative gradient. The supplied points satisfy this pattern. Gradient is (−5.12−5.12)/(0.080−(−0.080)) = −64.0 s⁻². Therefore ω = √64.0 = 8.00 rad s⁻¹. Period is T = 2π/ω = 0.785 s. Taking the largest measured displacement as amplitude gives v_max = ωA = (8.00)(0.080) = 0.640 m s⁻¹. The effective spring constant is k = mω² = (0.250)(64.0) = 16.0 N m⁻¹. Total energy is ½kA² = 0.0512 J. A constant acceleration zero offset adds the same value to every vertical reading. It preserves the gradient and inferred ω but shifts the line vertically so it no longer passes through the origin, creating an apparent nonzero acceleration at equilibrium.

Displacement x / m	Acceleration a / m s⁻²
-0.08	5.12
-0.04	2.56
0	0
0.04	-2.56
0.08	-5.12

Test a simple-harmonic acceleration model

What a complete response should do

Acceleration against displacement

Your self-assessment