Simple Harmonic Motion — Question 2

Question 2

Two speed measurements reveal the hidden amplitude

An ideal SHM particle has speed $0.80\,\mathrm{m/s}$ at displacement $0.060\,\mathrm{m}$ and speed $0.60\,\mathrm{m/s}$ at displacement $0.080\,\mathrm{m}$. Both displacements are measured from equilibrium.

1. Derive $v^2=\omega^2(A^2-x^2)$ from sinusoidal kinematics and infer $\omega$ and $A$.

2. Can the two speed measurements determine the phase or the direction of travel at either point? Explain.

Question 2 - Solution

Two speed measurements reveal the hidden amplitude

(a) Derivation and calculation

Eliminate the phase from $x=A\cos\psi$ and $v=-A\omega\sin\psi$:$$\frac{x^2}{A^2}+\frac{v^2}{A^2\omega^2}=1\quad\Rightarrow\quad v^2=\omega^2(A^2-x^2).$$$$\omega^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}=\frac{0.80^2-0.60^2}{0.080^2-0.060^2}=100\,\mathrm{s^{-2}}.$$$$\boxed{\omega=10\,\mathrm{rad/s}},\qquad A=\sqrt{0.060^2+\frac{0.80^2}{100}}=\boxed{0.100\,\mathrm m}.$$

(b) Result and interpretation

Each speed permits two signed velocities:$$x=0.060:\ v=\pm0.80,\qquad x=0.080:\ v=\pm0.60\quad(\mathrm{m/s}).$$$$\psi=\operatorname{atan2}\!\left(-\frac v{A\omega},\frac xA\right),\qquad\phi=\psi-\omega t\pmod{2\pi}.$$Neither velocity sign nor observation time is specified, so the initial phase is undetermined.

Physical check

Both states satisfy $x^2+v^2/\omega^2=0.0100\,\mathrm{m^2}$; inferred amplitude exceeds both measured displacements.