Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$T/4$

$T/8$

$T/12$

$T/2$

Step-by-Step Solution

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 .
Substitute Target Displacement: We need to find the time $t$ when the particle has travelled half of its amplitude, so we set $x = \frac{A}{2}$ . $\frac{A}{2} = A \sin(\omega t)$ $\sin(\omega t) = \frac{1}{2}$
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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