PHSX 220 Physics I (w/ Calculus) · Unit 4: Oscillations, waves, sound · on Final (Exam 4)
Ch 15.1–15.3 Simple harmonic motion
Study notes for Simple harmonic motion in PHSX 220 Physics I (w/ Calculus) at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- SHM happens whenever the restoring force is proportional to displacement: $F = -kx$ gives $a = -\dfrac{k}{m}x$. Solution $x(t) = A\cos(\omega t + \phi)$ with $\omega = \sqrt{k/m}$, period $T = 2\pi\sqrt{m/k}$, frequency $f = 1/T$.
- Velocity $v(t) = -A\omega\sin(\omega t + \phi)$, acceleration $a(t) = -A\omega^2\cos(\omega t+\phi) = -\omega^2 x$. Max speed $A\omega$ at the center; max acceleration $A\omega^2$ at the extremes (where $v = 0$).
- The period does not depend on amplitude (for an ideal spring) or on $g$ (a vertical spring just shifts the equilibrium point).
- Energy: $E = \tfrac12 kA^2 = \tfrac12 mv^2 + \tfrac12 kx^2$, sloshing between kinetic and potential. Speed at position $x$: $v = \omega\sqrt{A^2 - x^2}$.
- SHM is the shadow of uniform circular motion: the projection of a point moving on a circle of radius $A$ at angular speed $\omega$.
Key formulas
Motion
$$x = A\cos(\omega t+\phi),\quad \omega = \sqrt{\frac km},\quad T = 2\pi\sqrt{\frac mk},\quad v_{\max} = A\omega,\quad a_{\max} = A\omega^2$$
Energy
$$E = \tfrac12 kA^2 = \tfrac12 mv^2 + \tfrac12 kx^2,\qquad v = \omega\sqrt{A^2 - x^2}$$
Worked example
A 0.50 kg mass on a 200 N/m spring oscillates with amplitude 0.10 m. Find $T$, $v_{\max}$, and the speed at $x = 0.06$ m.
Show the solution
$\omega = \sqrt{400} = 20$ rad/s, $T = 2\pi/20 = 0.314$ s. $v_{\max} = 0.1(20) = 2.0$ m/s. $v = 20\sqrt{0.01 - 0.0036} = 1.6$ m/s.
Common mistakes
- Confusing ω (rad/s) with f (Hz); ω = 2πf.
- Thinking a larger amplitude changes the period.
- Using degrees in cos(ωt).
Exam tipTwo anchors: v is biggest at the center, a is biggest at the ends. Most conceptual questions reduce to that.
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