Oscillations and SHM: Application Problems (3)

Question

A block of mass $m = 0.5$ kg is attached to a spring of stiffness $k = 200$ N/m on a frictionless horizontal surface. The block is pulled $0.1$ m from equilibrium and released. Find (a) the time period, (b) the maximum speed, and (c) the speed when displacement is $0.06$ m.

Solution — Step by Step

Why This Works

In SHM, total energy is conserved: $\frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2$. Solving for $v$ gives $v = \omega \sqrt{A^2 - x^2}$ — a relation worth memorising.

The time period $T = 2\pi \sqrt{m/k}$ depends only on mass and spring constant, not on amplitude. This isochronism is the defining feature of SHM.

Alternative Method

Use the parametric form $x(t) = A \cos(\omega t)$, $v(t) = -A\omega \sin(\omega t)$. To find $v$ at a given $x$, solve for $\sin(\omega t) = \pm\sqrt{1 - (x/A)^2}$ and substitute. Same answer, more steps — energy conservation is cleaner.

Common Mistake

Students sometimes write $v = \omega \sqrt{x^2 - A^2}$, getting the sign wrong. Since $|x| \leq A$, that would give an imaginary number. The correct form is $v = \omega \sqrt{A^2 - x^2}$ — amplitude squared minus displacement squared.