Question

If the differential equation for a simple harmonic motion is \( \frac{d^2y}{dt^2} + 2y = 0 \), the time-period of the motion is

• A. \( \pi\sqrt{2} \text{ s} \)
• B. \( \frac{\sqrt{2}}{\pi} \text{ s} \)
• C. \( \frac{\pi}{\sqrt{2}} \text{ s} \)
• D. \( 2\pi \text{ s} \)
• E. \( \frac{\sqrt{\pi}}{2} \text{ s} \)

Hint:

Always check that the coefficient of the \( \frac{d^2y}{dt^2} \) term is 1 before identifying \( \omega^2 \).

Answer 1

The Correct Option is A

Concept: The motion of a particle executing Simple Harmonic Motion (SHM) is described by a specific second-order differential equation: \[ \frac{d^2y}{dt^2} + \omega^2 y = 0 \] Where:
• \( \omega \) is the angular frequency of the oscillation (in rad/s).
• \( T = \frac{2\pi}{\omega} \) is the time period, representing the time required for one full oscillation cycle.

Step 1: Identify the angular frequency from the given equation.
Comparing the given equation \( \frac{d^2y}{dt^2} + 2y = 0 \) to the standard form \( \frac{d^2y}{dt^2} + \omega^2 y = 0 \): \[ \omega^2 = 2 \implies \omega = \sqrt{2} \text{ rad/s} \]

Step 2: Solve for the Time Period \(T\).
Substitute the value of \( \omega \) into the time period formula: \[ T = \frac{2\pi}{\sqrt{2}} \] To simplify, multiply the numerator and denominator by \( \sqrt{2} \) (or realize \( 2 = \sqrt{2} \cdot \sqrt{2} \)): \[ T = \frac{\sqrt{2} \cdot \sqrt{2} \cdot \pi}{\sqrt{2}} = \sqrt{2}\pi \text{ or } \pi\sqrt{2} \text{ s} \]