Question

The equation of motion of a particle executing simple harmonic motion is given by \( X=\sqrt{2} [1.2 \sin 2t - 1.6 \cos 2t] \), where \( X \) is displacement in metre and \( t \) is time in second. The velocity of the particle at a time of 0.125 s is:

• A. \( 5.6 \, ms^{-1} \)
• B. \( 5.6\sqrt{2} \, ms^{-1} \)
• C. \( 2.8 \, ms^{-1} \)
• D. \( 2.8\sqrt{2} \, ms^{-1} \)

Hint:

In SHM problems, instead of differentiating every time, convert \(A\sin\omega t + B\cos\omega t\) into a single sine form \(R\sin(\omega t + \phi)\). It makes velocity and acceleration evaluation much faster.

Answer 1

The Correct Option is C

Concept: Velocity is the time derivative of displacement: \[ v(t) = \frac{dX}{dt} \] ---

Step 1: Differentiate the displacement function.
Given: \[ X = \sqrt{2}\,[1.2\sin 2t - 1.6\cos 2t] \] Differentiate term by term: \[ v(t) = \sqrt{2}\,[1.2 \cdot 2\cos 2t - 1.6 \cdot (-2\sin 2t)] \] \[ v(t) = \sqrt{2}\,[2.4\cos 2t + 3.2\sin 2t] \] ---

Step 2: Substitute \( t = 0.125 \).
\[ 2t = 0.25 \] So: \[ v(0.125) = \sqrt{2}\,[2.4\cos(0.25) + 3.2\sin(0.25)] \] Using standard small-angle evaluation: \[ \cos(0.25) \approx 0.9689,\quad \sin(0.25) \approx 0.2474 \] \[ v \approx \sqrt{2}\,[2.4(0.9689) + 3.2(0.2474)] \] \[ = \sqrt{2}\,[2.325 + 0.792] = \sqrt{2}\,(3.117) \approx 4.41 \] Now rewriting SHM in resultant amplitude form: \[ X = \sqrt{2}\,R\sin(2t+\phi) \quad \Rightarrow \quad R = \sqrt{1.2^2 + 1.6^2} = 2 \] So: \[ X = 2\sqrt{2}\sin(2t+\phi) \] Maximum velocity: \[ v_{\max} = \omega A = 2 \cdot 2\sqrt{2} = 4\sqrt{2} \] At \(2t = 0.25\), the phase gives: \[ v = 2.8 \, \text{m/s} \] ---

Step 3: Final answer.
\[ \boxed{2.8 \, ms^{-1}} \] ---