Question

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is \(16 \times 10^{-3} \mathrm{~J}\). The equation of motion of this particle is (Initial phase of oscillation \(=0^{\circ}\) )

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

Hint:

The kinetic energy at the mean position is the maximum, equal to half m omega squared A squared.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
At the mean position the kinetic energy is maximum: \(K=\dfrac{1}{2}m\omega^{2}A^{2}\).

Step 2: Substitute
\(m=0.2\text{ kg}\), \(A=0.2\text{ m}\), \(K=16\times10^{-3}\text{ J}\).
\(16\times10^{-3}=\dfrac{1}{2}(0.2)\omega^{2}(0.04)=0.004\,\omega^{2}\)

Step 3: Calculation
\(\omega^{2}=4\), so \(\omega=2\text{ rad/s}\).

Step 4: Equation
With zero initial phase, \(y=A\sin\omega t=0.2\sin(2t)\).

Final Answer:
The equation of motion is \(y=0.2\sin(2t)\). \[ \boxed{\text{(D) }\mathrm{Y}=0.2\sin(2t)} \]