Question

A particle is moving along the \(x\)-axis where the speed varies as \[ v^2 = 100 - x^2 \] Determine the time period.

• A. \(4\pi\)
• B. \(8\pi\)
• C. \(2\pi\)
• D. \(\pi\)

Answer 1

The Correct Option is C

Concept:
For simple harmonic motion (SHM), velocity as a function of displacement is \[ v = \omega\sqrt{A^2 - x^2} \] where \(A\) = amplitude, \(\omega\) = angular frequency. Step 1: Rewrite the given equation. \[ v^2 = 100 - x^2 \] \[ v = \sqrt{100 - x^2} \] Step 2: Compare with SHM velocity relation. \[ v = \omega\sqrt{A^2 - x^2} \] Comparing, \[ A^2 = 100 \] \[ A = 10 \] and \[ \omega = 1 \] Step 3: Find the time period. \[ T = \frac{2\pi}{\omega} \] \[ T = \frac{2\pi}{1} \] \[ T = 2\pi \]