Simple Harmonic Motion: Application Problems (5)

Question

A spring of force constant $k = 200$ N/m is attached to a $0.5$ kg block on a frictionless horizontal surface. The block is pulled $0.1$ m from equilibrium and released from rest. Find: (a) the angular frequency, (b) the time period, (c) the maximum speed, (d) the speed at $x = 0.05$ m.

Solution — Step by Step

Why This Works

In SHM, total mechanical energy is constant: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}m v_{max}^2$. At any displacement $x$, $\tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2 = \tfrac{1}{2}kA^2$, which rearranges to $v = \omega\sqrt{A^2 - x^2}$.

The time period $T = 2\pi\sqrt{m/k}$ depends only on $m$ and $k$ — not on amplitude. Push it more, and it just goes faster but takes the same time per cycle.

Alternative Method

Energy conservation directly:

$\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$

$v = \sqrt{\frac{k(A^2 - x^2)}{m}} = \sqrt{\frac{200(0.01 - 0.0025)}{0.5}} = \sqrt{3} \approx 1.73 \text{ m/s} \checkmark$

Common Mistake

The single biggest error: confusing $A$ (amplitude) with the instantaneous displacement. Amplitude is the maximum displacement, fixed for a given motion. Instantaneous $x$ varies between $-A$ and $+A$.

Also: students sometimes write $v = \omega A$ at every point, which is wrong. That’s only the maximum speed (at $x = 0$). Use $v = \omega\sqrt{A^2 - x^2}$ for general $x$.