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: Approach
Use energy conservation.

Step 2: Energy
Maximum speed is $v_{max}=\dfrac{2\pi a}{T}$. Kinetic energy at $x=a/3$ is $1-\dfrac19=\dfrac89$ of the maximum.

Step 3: Speed
$v=v_{max}\sqrt{\dfrac89}=\dfrac{2\pi a}{T}\cdot\dfrac{2\sqrt2}{3}=\dfrac{4\sqrt2\pi a}{3T}$.

Step 4: Answer
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). \[ \boxed{\frac{4\sqrt2\pi a}{3T}} \]