Question

A particle executes S.H.M. starting from the mean position. Its amplitude is ' a ' and its periodic time is ' T '. At a certain instant, its speed ' u ' is half that of maximum speed \( V_{\text{max}} \). The displacement of the particle at that instant is

• A. \( \frac{2a}{\sqrt{3}} \)
• B. \( \frac{\sqrt{2}a}{3} \)
• C. \( \frac{3a}{\sqrt{2}} \)
• D. \( \frac{\sqrt{3}a}{2} \)

Hint:

At $x = \frac{\sqrt{3}}{2}a$, velocity is half of max. At $x = \frac{1}{2}a$, velocity is $\frac{\sqrt{3}}{2}$ of max.

Answer 1

The Correct Option is D

Step 1: Formula for Velocity
$v = \omega \sqrt{a^2 - x^2}$. Maximum speed $V_{max} = a\omega$.
Step 2: Given Condition
$v = \frac{1}{2} V_{max} \implies \omega \sqrt{a^2 - x^2} = \frac{1}{2} a\omega$.
Step 3: Solve for x
$\sqrt{a^2 - x^2} = \frac{a}{2} \implies a^2 - x^2 = \frac{a^2}{4}$.
$x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4} \implies x = \frac{\sqrt{3}a}{2}$.
Final Answer: (D)