Question

A particle is performing S.H.M. about \(x = 0\), with an amplitude \(a\) and time period \(T\). The speed of the particle at \(x = \frac{a}{3}\) will be

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

Hint:

Speed in SHM is omega times root of (a squared minus x squared).

Answer 1

The Correct Option is D

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

Step 2: Substitute
\[ v=\frac{2\pi}{T}\sqrt{a^2-\frac{a^2}{9}}=\frac{2\pi}{T}\cdot a\sqrt{\frac89} \]
\[ \sqrt{\frac89}=\frac{2\sqrt2}{3} \]

Step 3: Result
\[ v=\frac{2\pi}{T}\cdot\frac{2\sqrt2a}{3}=\frac{4\sqrt2\pi a}{3T} \]

Step 4: Check the options
The option \(\frac{2\pi a}{T}\) is the maximum speed at the mean position. The others have a different numerical factor. The answer is option (D).

Final Answer:
The speed is omega times root of (a squared - a squared/9), which is 4 root 2 pi a over 3T, option (D). \[ \boxed{\frac{4\sqrt2\pi a}{3T}} \]