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(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $A = 10 \,cm$ and $x = 5 \

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$1 \,m/s$

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(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $A = 10 \,cm$ and $x = 5 \,cm$,we have $5 = 10 \sin(\omega t)$,which implies $\sin(\omega t) = 1/2$.Thus,$\omega t = \pi/6$. Since $\omega = 2\pi/T$,we get $t = \frac{\pi/6}{2\pi/T} = T/12$.Given $T = 0.6 \,s$,the time interval is $t = 0.6/12 = 0.05 \,s$.The mean velocity $V_{	ext{mean}}$ is defined as the total displacement divided by the total time interval:$V_{	ext{mean}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$.Starting from equilibrium ($x_i = 0$ at $t_i = 0$) to $x_f = 5 \,cm$ at $t_f = 0.05 \,s$:$V_{	ext{mean}} = \frac{5 \,cm - 0}{0.05 \,s} = 100 \,cm/s = 1 \,m/s$.

$A$ particle executes simple harmonic motion with a time period $0.6 \,s$ and amplitude $10 \,cm$. Find the mean velocity of the particle over the time interval during which it travels a distance $5 \,cm$ starting from the equilibrium position.

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