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.
| Quantity | Equation |
|---|---|
| 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:
- 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.
- 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
- For a mass $m=0.3\,\text{kg}$ on a spring with $k=12\,\text{N/m}$, calculate the angular frequency $\omega$.
- If the amplitude $A$ is $0.05\,\text{m}$ and $\omega = 4\,\text{rad/s}$, what is the maximum speed $v_{\text{max}}$?
- 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!
Revision
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