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PhysicsClassical Physics

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.