Question

The angular displacement of a body performing circular motion is given by $\theta = 5 \sin\left(\frac{\pi t}{6}\right)$. The angular velocity of the body at $t = 3\ \text{second}$ will be [$\sin\left(\frac{\pi}{2}\right) = 1$, $\cos\left(\frac{\pi}{2}\right) = 0$]

• A. $5\ \text{rad/s}$
• B. $1\ \text{rad/s}$
• C. $2.5\ \text{rad/s}$
• D. $\text{zero}\ \text{rad/s}$

Hint:

Think about the physical nature of harmonic oscillations! A sine wave displacement function reaches its maximum peak value when its internal angle equals $\frac{\pi}{2}$. At $t=3$, the angle is $\frac{3\pi}{6} = \frac{\pi}{2}$, meaning the object has reached its extreme turning point. Since an object momentarily stops to change direction at any extreme turning point, its velocity must equal zero!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Determine the velocity of a particle executing simple harmonic motion at a specific time using the phase of its displacement function.

Step 2: Key Formula or Approach:
For displacement x = A sin(ωt), velocity v = Aω cos(ωt). When the phase angle ωt equals π/2, the sine function reaches its peak (x = A) and the cosine function drops to zero.

Step 3: Detailed Explanation:
At t = 3, the phase angle is 3π/6 = π/2. This corresponds to the extreme turning point of the oscillation, where the particle momentarily halts before reversing direction. At any turning point, displacement is maximized and velocity is instantaneously zero. Recognizing that π/2 signifies an amplitude extremum provides an immediate physical answer without evaluating the cosine function.

Step 4: Final Answer:
The velocity at that instant is zero.