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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
A
$\dfrac{T}{8}$
B
$\dfrac{T}{12}$
C
$\dfrac{T}{2}$
D
$\dfrac{T}{4}$
Detailed Solution
Starting from the equilibrium position, the displacement is $y = a\sin\left(\dfrac{2\pi}{T}\right)t$
According to the problem, $y = \dfrac{a}{2}$
$\dfrac{a}{2} = a\sin\left(\dfrac{2\pi}{T}\right)t$
$\sin\left(\dfrac{2\pi}{T}\right)t = \dfrac{1}{2}$
$\Rightarrow \dfrac{\pi}{6} = \left(\dfrac{2\pi}{T}\right)t$
$\Rightarrow t = \dfrac{T}{12}$
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.
According to the problem, $y = \dfrac{a}{2}$
$\dfrac{a}{2} = a\sin\left(\dfrac{2\pi}{T}\right)t$
$\sin\left(\dfrac{2\pi}{T}\right)t = \dfrac{1}{2}$
$\Rightarrow \dfrac{\pi}{6} = \left(\dfrac{2\pi}{T}\right)t$
$\Rightarrow t = \dfrac{T}{12}$
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.