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NEET 2007 (AIPMT) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from NEET 2007 (AIPMT)

Q74NEET 2007 · AIPMTEasy

Periodic Motion & SHM Basics

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

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Answer: (b)

From the mean position: $x = a\sin\frac{2\pi t}{T}$ .
$\frac{a}{2} = a\sin\frac{2\pi t}{T} \Rightarrow \frac{2\pi t}{T} = \frac{\pi}{6}$ .
$t = \frac{T}{12}$ .
So the answer is (b) $T/12$ .

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