A particle executes SHM of amplitude 25, cm a | vedclass

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Question

(A) The equation of motion for a particle in $SHM$ starting from the mean position is given by $y = A \sin(\omega t)$,where $A = 25\, cm$ and $\omega

Vedclass · Accepted Answer

$0.5$

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(A) The equation of motion for a particle in $SHM$ starting from the mean position is given by $y = A \sin(\omega t)$,where $A = 25\, cm$ and $\omega = \frac{2\pi}{T} = \frac{2\pi}{3}$.To find the time taken to reach $y = 12.5\, cm$ from the mean position $(y = 0)$:$12.5 = 25 \sin(\frac{2\pi}{3} t)$$\frac{1}{2} = \sin(\frac{2\pi}{3} t)$Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$,we have $\frac{\pi}{6} = \frac{2\pi}{3} t$.Solving for $t$,we get $t = \frac{1}{4} = 0.25\, s$.The particle moves from $-12.5\, cm$ to $+12.5\, cm$ passing through the mean position. The total time taken is $2t = 2 	imes 0.25 = 0.5\, s$.

$A$ particle executes $SHM$ of amplitude $25\, cm$ and time period $3\, s$. What is the minimum time required for the particle to move between two points $12.5\, cm$ on either side of the mean position?

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