Question

A particle is performing SHM of amplitude \(A\). Find the time required for particle to go from mean position to \( \dfrac{A}{\sqrt{2}} \). Time period of SHM is \(5\) sec :-

• A. \( \dfrac{5}{4} \) sec.
• B. \( \dfrac{5}{12} \) sec.
• C. \( \dfrac{5}{8} \) sec.
• D. \( \dfrac{5}{6} \) sec.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The displacement of a particle executing Simple Harmonic Motion (SHM) starting from the mean position is represented by a sine function. We find the phase angle corresponding to the given displacement to calculate the time taken.
Step 2: Key Formula or Approach:
The equation of SHM starting from the mean position is: \[ x(t) = A \sin(\omega t) \] where \(\omega = \frac{2\pi}{T}\).
Step 3: Detailed Explanation:
The particle travels to \(x = \frac{A}{\sqrt{2}}\). Substitute this into the displacement equation: \[ \frac{A}{\sqrt{2}} = A \sin(\omega t) \] \[ \sin(\omega t) = \frac{1}{\sqrt{2}} \] The smallest positive angle is \(\omega t = \frac{\pi}{4}\) radians. Substitute \(\omega = \frac{2\pi}{T}\): \[ \left(\frac{2\pi}{T}\right) t = \frac{\pi}{4} \] \[ \frac{2t}{T} = \frac{1}{4} \implies t = \frac{T}{8} \] Given the time period \(T = 5 \, \text{secs}\): \[ t = \frac{5}{8} \, \text{secs} \]
Step 4: Final Answer:
The time required is \(5/8 \, \text{sec.}\)