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: Understanding the Concept:
Total energy is \(\tfrac12kA^2\) and potential energy at displacement \(x\) is \(\tfrac12kx^2\). Kinetic energy is half the total when the potential energy is also half.

Step 2: Find the displacement:
\(\tfrac12kx^2=\tfrac12\cdot\tfrac12kA^2\) gives \(x=\dfrac A{\sqrt2}\).

Step 3: Use x = A sin(omega t):
The particle starts from the mean position, so \(x=A\sin\omega t\). Thus \(\sin\omega t=\dfrac1{\sqrt2}\), so \(\omega t=\dfrac\pi4\).

Step 4: Find time:
\(\omega=\dfrac{2\pi}T\), so \(t=\dfrac T8=\dfrac68=0.75\) s.

Step 5: Choose:
Option (A).

Final Answer:
The kinetic energy is 50 percent at t = T/8 = 0.75 s. \[ \boxed{0.75\ \text{s}} \]