Question

An object undergoing simple harmonic motion takes \(0.5 \, s\) to travel from one point of zero velocity to the next such point. The angular frequency of the motion is

• A. \(\pi \, rad \, s^{-1}\)
• B. \(2\pi \, rad \, s^{-1}\)
• C. \(3\pi \, rad \, s^{-1}\)
• D. \(\dfrac{\pi}{2} \, rad \, s^{-1}\)

Hint:

In SHM, the time taken to move from one extreme position to the opposite extreme position is always equal to \(\dfrac{T}{2}\).

Answer 1

The Correct Option is B

Step 1: Identify the points where velocity becomes zero.
In simple harmonic motion, velocity becomes zero at the extreme positions.
The object moves from one extreme position to the opposite extreme position in half of the time period.
Therefore,
\[ \frac{T}{2}=0.5 \, s \]

Step 2: Calculate the time period.
\[ T=2\times0.5 \] \[ T=1 \, s \]

Step 3: Use the relation between angular frequency and time period.
Angular frequency is given by
\[ \omega=\frac{2\pi}{T} \] Substituting \(T=1 \, s\),
\[ \omega=\frac{2\pi}{1} \] \[ \omega=2\pi \, rad \, s^{-1} \]

Step 4: Final conclusion.
Hence, the angular frequency of the motion is
\[ \boxed{2\pi \, rad \, s^{-1}} \]