A particle executes a simple harmonic motion | vedclass

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(D) The displacement of a particle in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$.Here,$A$ is the amplit

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$T / 12$

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(D) The displacement of a particle in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$.Here,$A$ is the amplitude,$\omega = \frac{2\pi}{T}$ is the angular frequency,and $T$ is the time period.We want to find the time $t$ when the displacement $y = \frac{A}{2}$.Substituting the values into the equation: $\frac{A}{2} = A \sin\left(\frac{2\pi t}{T}\right)$.Dividing both sides by $A$: $\frac{1}{2} = \sin\left(\frac{2\pi t}{T}\right)$.Since $\sin(30^\circ) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\frac{2\pi t}{T} = \frac{\pi}{6}$.Solving for $t$: $t = \frac{T}{12}$.

$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

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