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? (\(cos45^{\circ} = 1/\sqrt{2}\))

• A. \(0.75\) s
• B. \(0.50\) s
• C. \(0.25\) s
• D. \(3\) s

Hint:

KE is half when x = A/sqrt2, which is a phase of 45 degrees.

Answer 1

The Correct Option is A

Step 1: Use KE formula:
$K=\tfrac12m\omega^2A^2\cos^2\omega t$ for a start at the mean position.

Step 2: Set 50 percent:
$\cos^2\omega t=\tfrac12$, so $\cos\omega t=\dfrac1{\sqrt2}$ and $\omega t=45^\circ=\dfrac\pi4$.

Step 3: Time:
$t=\dfrac{\pi/4}{2\pi/6}=\dfrac68=0.75$ s. Option (A).

Final Answer:
The phase is 45 degrees, which is T/8. \[ \boxed{A} \]