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: Understanding the Question:
The kinetic energy at the mean position is the maximum kinetic energy. We can use this to find the angular frequency \(\omega\). With amplitude \(A\) and initial phase \(\phi = 0\), we can write the S.H.M. equation \(Y = A \sin(\omega t + \phi)\).
Step 2: Key Formula or Approach:
Maximum Kinetic Energy: \(KE_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m \omega^2 A^2\).
Step 3: Detailed Explanation:
Given:
Mass \(m = 200 \text{ g} = 0.2 \text{ kg}\)
Amplitude \(A = 0.2 \text{ m}\)
\(KE_{\text{max}} = 16 \times 10^{-3} \text{ J}\)
Initial phase \(\phi = 0^{\circ}\)
Equating:
\[ 16 \times 10^{-3} = \frac{1}{2} \times 0.2 \times \omega^2 \times (0.2)^2 \]
\[ 16 \times 10^{-3} = 0.1 \times \omega^2 \times 0.04 \]
\[ 16 \times 10^{-3} = 0.004 \times \omega^2 \]
\[ \omega^2 = \frac{16 \times 10^{-3}}{4 \times 10^{-3}} = 4 \]
\[ \omega = 2 \text{ rad/s} \]
The displacement equation is \(Y = A \sin(\omega t)\):
\[ Y = 0.2 \sin(2t) \]
Step 4: Final Answer:
The equation of motion is \(Y = 0.2 \sin(2t)\).