A particle executing S.H.M. with an amplitude | vedclass

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(C) The equation of motion for a particle starting from the positive extreme position is given by $y = a \cos(\omega t)$.Here,$a$ is the amplitude and

Vedclass · Accepted Answer

$0.67$

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(C) The equation of motion for a particle starting from the positive extreme position is given by $y = a \cos(\omega t)$.Here,$a$ is the amplitude and $\omega$ is the angular frequency.We are given $a = 4 \, cm$ and $T = 4 \, s$.We need to find the time $t$ when the displacement $y = \frac{a}{2}$.Substituting the values into the equation: $\frac{a}{2} = a \cos(\omega t)$.This simplifies to $\cos(\omega t) = \frac{1}{2}$.Therefore,$\omega t = \frac{\pi}{3}$.Since $\omega = \frac{2\pi}{T}$,we have $\frac{2\pi t}{T} = \frac{\pi}{3}$.Solving for $t$: $t = \frac{T}{6}$.Given $T = 4 \, s$,we get $t = \frac{4}{6} = \frac{2}{3} \, s \approx 0.67 \, s$.

$A$ particle executing $S.H.M.$ with an amplitude of $4 \, cm$ and a time period $T = 4 \, s$. The time taken by it to move from the positive extreme position to half the amplitude is ..... $s$.

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