Question

A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is $50\%$ of total energy? \( (\cos 45^\circ = \frac{1}{\sqrt{2}}) \)

• A. 0.25 s
• B. 0.50 s
• C. 0.75 s
• D. 1 s

Hint:

- $K/E = \cos^2(\omega t)$ (starting from mean) - Use $T = \frac{2\pi}{\omega}$

Answer 1

The Correct Option is C

Concept: \[ \frac{K}{E} = \cos^2(\omega t) \]

Step 1: Given condition.
\[ \frac{K}{E} = \frac{1}{2} \Rightarrow \cos^2(\omega t) = \frac{1}{2} \Rightarrow \cos(\omega t) = \frac{1}{\sqrt{2}} \]

Step 2: Find angle.
\[ \omega t = 45^\circ = \frac{\pi}{4} \]

Step 3: Find $\omega$.
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \]

Step 4: Solve for time.
\[ \frac{\pi}{3} t = \frac{\pi}{4} \Rightarrow t = \frac{3}{4} = 0.75\ \text{s} \]