A particle executes simple harmonic oscillation with an amplitude $A$. The period of oscillation is $T$. The m

Question

A particle executes simple harmonic oscillation with an amplitude $A$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

Accepted Answer

1. **Identify Equation of Motion:** For a particle starting its simple harmonic motion from the equilibrium (mean) position, its displacement $x$ as a function of time $t$ is given by $x = A \sin(\omega t)$, where $A$ is the amplitude and $\omega$ is the angular frequency .
2. **Substitute Target Displacement:** We need to find the time $t$ when the particle has travelled half of its amplitude, so we set $x = \frac{A}{2}$.
   $$ \frac{A}{2} = A \sin(\omega t) $$
   $$ \sin(\omega t) = \frac{1}{2} $$
3. **Solve for Time ($t$):** The minimum time corresponds to the smallest positive angle whose sine is $\frac{1}{2}$, which is $\frac{\pi}{6}$ radians ($30^\circ$).
   $$ \omega t = \frac{\pi}{6} $$
   We know that angular frequency $\omega = \frac{2\pi}{T}$ . Substituting this into the equation:
   $$ \left(\frac{2\pi}{T}\right) t = \frac{\pi}{6} $$
   $$ \frac{2t}{T} = \frac{1}{6} $$
   $$ t = \frac{T}{12} $$

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