Oscillations and SHM: Speed-Solving Techniques (6)

Question

A particle executes SHM with amplitude $A = 10$ cm and time period $T = 2$ s. At $t = 0$, the particle is at $x = 5$ cm and moving in the positive direction. Find (a) its velocity at $x = 8$ cm, and (b) the time it takes to reach $x = 10$ cm for the first time.

This is a classic JEE Main pattern that rewards using the SHM “circle of reference” instead of plugging into formulas blindly.

Solution — Step by Step

Final answers: (a) $v = 6\pi \approx 18.85$ cm/s, (b) $t = 1/3$ s.

Why This Works

SHM is the projection of uniform circular motion onto a diameter. The radius of the reference circle is $A$, and the angular speed is $\omega$. At any point, the particle’s $x$-coordinate is $A\sin(\omega t + \phi)$. The phase $\phi$ tells us where on the circle we start.

The shortcut $v = \omega\sqrt{A^2 - x^2}$ comes from energy conservation: $\tfrac{1}{2}m\omega^2 A^2 = \tfrac{1}{2}m\omega^2 x^2 + \tfrac{1}{2}m v^2$. No need to take a derivative.

Alternative Method

Use the reference circle directly. At $t = 0$, the rotating vector points at angle $\pi/6$ from the $x$-axis (since $\sin(\pi/6) = 0.5$). To reach $x = A$, the vector must point straight up at angle $\pi/2$. The angle to sweep is $\pi/2 - \pi/6 = \pi/3$. Time = angle/$\omega$ = $(\pi/3)/\pi = 1/3$ s. Faster than the algebra route.

Common Mistake

Students often write $x(t) = A\cos(\omega t)$ by default, then get stuck when initial conditions disagree. Use $x(t) = A\sin(\omega t + \phi)$ and solve for $\phi$ from the given initial position and velocity sign. That handles all cases cleanly.