A particle is performing SHM with a path leng | vedclass

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Question

(A) The path length (total amplitude range) is $2A = 8 \ cm$.Therefore,the amplitude $A = 4 \ cm$.The time taken to travel from the mean position to t

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$2 \pi \ cm/s$

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(A) The path length (total amplitude range) is $2A = 8 \ cm$.Therefore,the amplitude $A = 4 \ cm$.The time taken to travel from the mean position to the extreme position is one-fourth of the time period $T$,which is given as $1 \ s$.So,$T/4 = 1 \ s$,which implies $T = 4 \ s$.The velocity at the equilibrium (mean) position is the maximum velocity,given by $v_{\max} = A\omega$.Substituting $\omega = 2\pi/T$,we get $v_{\max} = A(2\pi/T)$.$v_{\max} = 4 	imes (2\pi / 4) = 2\pi \ cm/s$.

$A$ particle is performing $SHM$ with a path length of $8 \ cm$. If it takes $1 \ s$ to reach from the mean position to the extreme position,what is its velocity at the equilibrium position?

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