use the equations v = v0 cos ωt and v = ± ω ()xx022−

Simple Harmonic Oscillations (SHO)

Think of a playground swing or a mass attached to a spring. When you push it and let go, it moves back and forth in a regular pattern. That regular back‑and‑forth motion is called simple harmonic motion (SHM).

Key Equations

In SHM we use a few handy equations. They let us predict where the object will be and how fast it’s moving.

QuantityEquation
Displacement\$x(t)=A\cos(\omega t+\phi)\$
Velocity\$v(t)=-A\omega\sin(\omega t+\phi)\$
Acceleration\$a(t)=-\omega^2x(t)\$
Energy\$E=\tfrac{1}{2}kA^2\$

Here, \$A\$ is the maximum displacement (amplitude), \$\omega\$ is the angular frequency, \$k\$ is the spring constant, and \$\phi\$ is the phase shift.

Velocity in SHM

Two common ways to write the velocity:

  1. Using the initial velocity \$v_0\$:

    \$v(t)=v_0\cos(\omega t)\$

    This form is handy when you know the speed at the start of the motion.

  2. Using the displacement \$x\$:

    \$v=\pm\omega\sqrt{A^2-x^2}\$

    The “±” shows that the speed is the same whether the object is moving right or left.

Both equations give the same result; they’re just different ways of looking at the same physics.

Example: Mass‑Spring System

Imagine a mass \$m=0.5\,\text{kg}\$ attached to a spring with \$k=20\,\text{N/m}\$. Pull the mass 0.1 m to the right and let go.

  • Amplitude \$A = 0.1\,\text{m}\$.
  • Angular frequency \$\omega = \sqrt{k/m} = \sqrt{20/0.5} \approx 6.32\,\text{rad/s}\$.
  • Maximum speed \$v_{\text{max}} = A\omega \approx 0.632\,\text{m/s}\$.

At the instant the mass passes the equilibrium point (where \$x=0\$), its speed is at the maximum, \$v_{\text{max}}\$.

Analogy: The Swing 🎠

Picture a child on a swing. When the child pushes off the ground, the swing goes forward and then swings back. The motion is almost perfectly SHM if the swing is not too damped.

  • Amplitude = how far the swing swings from the middle.
  • Angular frequency = how fast the swing completes a cycle.
  • Velocity is fastest at the middle of the swing and zero at the farthest points.

Just like a spring, the swing’s motion can be described by the same equations!

Quick Practice Problems

  1. For a mass \$m=0.3\,\text{kg}\$ on a spring with \$k=12\,\text{N/m}\$, calculate the angular frequency \$\omega\$.
  2. If the amplitude \$A\$ is \$0.05\,\text{m}\$ and \$\omega = 4\,\text{rad/s}\$, what is the maximum speed \$v_{\text{max}}\$?
  3. At \$t=0\$, a mass is at the equilibrium point moving to the right with speed \$0.4\,\text{m/s}\$. If \$\omega=5\,\text{rad/s}\$, find the amplitude \$A\$.

Try solving them before checking the solutions in the next lesson!