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: Use the maximum speed:
$v_{max} = \omega a = \frac{2\pi a}{T}$ and $\frac{v}{v_{max}} = \sqrt{1 - \frac{x^2}{a^2}}$.

Step 2: Evaluate:
$\sqrt{1 - \frac19} = \frac{2\sqrt2}{3}$, so $v = \frac{2\pi a}{T}\cdot\frac{2\sqrt2}{3} = \frac{4\sqrt2\pi a}{3T}$.

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