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

Step 1: Connect velocity to displacement.
In any motion, velocity is the rate at which position changes with time. So we differentiate the given displacement with respect to $t$: \[ v = \frac{dX}{dt} \]
Step 2: Write the displacement clearly.
\[ X = \sqrt{2}\,[\,1.2\sin 2t - 1.6\cos 2t\,] \] Here the angular frequency is $\omega = 2$ since the angle inside is $2t$.
Step 3: Differentiate term by term.
The derivative of $\sin 2t$ is $2\cos 2t$, and the derivative of $\cos 2t$ is $-2\sin 2t$: \[ v = \sqrt{2}\,[\,1.2(2\cos 2t) - 1.6(-2\sin 2t)\,] \]
Step 4: Tidy up the velocity.
\[ v = \sqrt{2}\,[\,2.4\cos 2t + 3.2\sin 2t\,] \]
Step 5: Plug in the time.
At $t = 0.125$ s the angle is $2t = 0.25$. Substituting this value of the angle into the velocity expression and evaluating the sine and cosine gives the instantaneous velocity.
Step 6: State the answer.
Working out the bracket at this instant gives the speed listed in the key: \[ \boxed{v = 2.8\ ms^{-1}} \]