Question

A particle performs simple harmonic motion with period of 3 second. The time taken by it to cover a distance equal to half the amplitude from mean position is

• A. \( \frac{1}{4} \, \text{s} \)
• B. \( \frac{3}{4} \, \text{s} \)
• C. \( \frac{3}{2} \, \text{s} \)
• D. \( \frac{1}{2} \, \text{s} \)

Hint:

In SHM, to find the time taken to cover a certain distance, use the equation \( x = A \sin(\omega t) \) and solve for \( t \).

Answer 1

The Correct Option is A

Step 1: SHM formula.
In simple harmonic motion, the distance \( x \) from the mean position at any time \( t \) is given by: \[ x = A \sin(\omega t) \] where \( A \) is the amplitude and \( \omega = \frac{2\pi}{T} \) is the angular frequency, and \( T \) is the period.
Step 2: Time for half the amplitude.
For half the amplitude, \( x = \frac{A}{2} \). So, we need to find \( t \) such that: \[ \frac{A}{2} = A \sin(\omega t) \] \[ \sin(\omega t) = \frac{1}{2} \] This gives: \[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] \[ t = \frac{\pi}{6\omega} = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12} \]
Step 3: Conclusion.
Since \( T = 3 \, \text{s} \), the time taken to cover half the amplitude is: \[ t = \frac{3}{12} = \frac{1}{4} \, \text{s} \] Thus, the correct answer is (A).