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  4. 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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Q. 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

NTA AbhyasNTA Abhyas 2022

Solution:

Using the equation of SHM,
$x\left(t\right)=Asin\left(\omega t\right)$ (from the equilibruim position)
At $x\left(t\right)=A/2$
$\therefore \frac{A}{2}=Asin\left(\omega t\right)$
$sin\left(\frac{\pi }{6}\right)=sin\left(\omega t\right)\text{or,}\frac{\pi }{6}=\frac{2 \pi t}{T}\left[\because \omega = \frac{2 \pi }{T}\right]$
or, $t=T/12$ .