Question

A particle of mass \(200 \text{ g}\) is executing S.H.M. of amplitude \(0.2 \text{ m}\). At mean position, K.E. is \(16 \times 10^{-3} \text{ J}\). The equation of motion is:

• A. \(Y = 0.2 \sin(4t)\)
• B. \(Y = 0.2 \sin(\frac{t}{4})\)
• C. \(Y = 0.2 \sin(\frac{t}{2})\)
• D. \(Y = 0.2 \sin(2t)\)

Hint:

The maximum velocity in S.H.M. occurs at the mean position and is equal to $\omega A$.

Answer 1

The Correct Option is D

Step 1: Find Angular Frequency
At mean position, \(K.E._{max} = \frac{1}{2}m\omega^2 A^2\).
\(16 \times 10^{-3} = \frac{1}{2} (0.2) \omega^2 (0.2)^2\).

Step 2: Calculation
\(0.016 = 0.1 \times \omega^2 \times 0.04 = 0.004 \omega^2\).
\(\omega^2 = \frac{0.016}{0.004} = 4 \implies \omega = 2 \text{ rad/s}\).

Step 3: Equation
\(Y = A \sin(\omega t) = 0.2 \sin(2t)\).
Final Answer: (D)