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(C) The given equation of motion is $X = 7 \cos(0.5 \pi t)$.Comparing this with the standard $SHM$ equation $X = A \cos(\omega t)$,we get the angular

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(C) The given equation of motion is $X = 7 \cos(0.5 \pi t)$.Comparing this with the standard $SHM$ equation $X = A \cos(\omega t)$,we get the angular frequency $\omega = 0.5 \pi \ rad/s$.The time period $T$ is given by $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{0.5 \pi} = 4 \ s$.The time taken by a particle to move from the equilibrium position (mean position) to the maximum displacement (extreme position) is one-fourth of the time period.Therefore,$t = \frac{T}{4} = \frac{4}{4} = 1 \ s$.

$A$ particle is oscillating according to the equation $X = 7 \cos(0.5 \pi t)$,where $t$ is in seconds. The particle moves from the position of equilibrium to maximum displacement in time ..... $s$.

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