Question

A particle is performing simple harmonic motion about \(x = 0\) with an amplitude '\(a\)' and periodic time T. The speed of the particle at \(x = \frac{a}{3}\) will be

• A. \(\frac{2πa}{T}\)
• B. \(\frac{4πa}{3T}\)
• C. \(\frac{4\sqrt{2}\,πa}{3T}\)
• D. \(\frac{\sqrt{3}\,π^2a}{2T}\)

Hint:

Use v = omega times the square root of (a squared minus x squared).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In SHM, the speed at displacement \(x\) is \(v = \omega\sqrt{a^2 - x^2}\), with \(\omega = \frac{2\pi}{T}\).

Step 2: Key Formula or Approach:
Put \(x = \frac a3\).

Step 3: Detailed Explanation:
\(a^2 - x^2 = a^2 - \frac{a^2}{9} = \frac{8a^2}{9}\), so \(\sqrt{a^2 - x^2} = \frac{2\sqrt2\,a}{3}\).
\[ v = \frac{2\pi}{T}\times\frac{2\sqrt2\,a}{3} = \frac{4\sqrt2\,\pi a}{3T} \]
Option A, \(\frac{2\pi a}{T}\), is the maximum speed at the mean position.

Final Answer:
The speed is \(\frac{4\sqrt{2}\pi a}{3T}\), option (C). \[ \boxed{\frac{4\sqrt{2}\,\pi a}{3T}} \]