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What Is Simple Harmonic Motion? Derive x(t), v(t), and a(t)

The oscillator has angular frequency 4.00 rad/s, with x(t) = 0.0200 cos(4.00t) m, v(t) = −0.0800 sin(4.00t) m/s, and a(t) = −0.320 cos(4.00t) m/s².

Question

A 2.00 kg2.00\,\mathrm{kg} block rests on a frictionless horizontal surface and is attached to a spring with force constant k=32.00 N/mk=32.00\,\mathrm{N/m}. The equilibrium position is x=0x=0. The block is pulled to x=+0.020 mx=+0.020\,\mathrm{m} and released from rest at t=0t=0. It oscillates between +0.020 m+0.020\,\mathrm{m} and −0.020 m-0.020\,\mathrm{m} with period T=1.57 sT=1.57\,\mathrm{s}.

Determine the position, velocity, and acceleration as functions of time.

Answer

Simple harmonic motion occurs when acceleration is proportional to displacement and points back toward equilibrium:

a=−ω2x.a=-\omega^2x.

The angular frequency is

ω=2πT=2π1.57 s=4.00 rad/s.\omega=\frac{2\pi}{T} =\frac{2\pi}{1.57\,\mathrm{s}} =4.00\,\mathrm{rad/s}.

The mass starts at x=+Ax=+A with zero velocity, so A=0.0200 mA=0.0200\,\mathrm{m} and ϕ=0\phi=0. The position is

x(t)=(0.0200 m)cos⁡(4.00t).x(t)=(0.0200\,\mathrm{m})\cos(4.00t).

Differentiating gives

v(t)=−(0.0800 m/s)sin⁡(4.00t)v(t)=-(0.0800\,\mathrm{m/s})\sin(4.00t)

and

a(t)=−(0.320 m/s2)cos⁡(4.00t).a(t)=-(0.320\,\mathrm{m/s^2})\cos(4.00t).

The spring parameters independently give

km=32.002.00=4.00 rad/s,\sqrt{\frac{k}{m}}=\sqrt{\frac{32.00}{2.00}}=4.00\,\mathrm{rad/s},

consistent with the stated period.

Evidence boundary

The equations describe an ideal undamped horizontal mass-spring oscillator obeying Hooke’s law, with t = 0 at maximum positive displacement. They use the stated period, which is consistent with sqrt(k/m) to the given precision. Damping, driving, spring mass, friction, and nonlinear motion are excluded.

Sources

These references support the concepts and methods used in the explanation above.

What Is Simple Harmonic Motion? Spring Equations | Verla