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A $5\; kg$ collar is attached to a spring of spring constant $500\;N m ^{-1} .$ It slides without friction over a hortzontal rod. The collar is displaced from its equilibrium position by $10.0\; cm$ and released. Calculate
$(a)$ the period of oscillation.
$(b)$ the maximum speed and
$(c)$ maximum acceleration of the collar.
Solution
$(a)$ The period of oscillation
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{5.0\, kg }{500\,N\,m^{-1}}}$
$=(2 \pi / 10)\, s$
$=0.63 \,s$
$(b)$ The velocity of the collar executing $SHM$ is given by
$v(t)=-A \omega \sin (\omega t+\phi)$
The maximum speed is given by,
$v_{m}=A \omega$
$=0.1 \times \sqrt{\frac{k}{m}}$
$=0.1 \times \sqrt{\frac{500\, N m ^{-1}}{5\, kg }}$
$=1 \,m s ^{-1}$
and it occurs at $x=0$
$(c)$ The acceleration of the collar at the displacement $x(t)$ from the equilibrium is given by,
$a(t) =-\omega^{2} x(t)$
$=-\frac{k}{m} x(t)$
Therefore, the maximum acceleration is $a_{\max }=\omega^{2} A$
$=\frac{500\, N \,m ^{-1}}{5 \,kg } \times 0.1 \,m$
$=10\, m s ^{-2}$
and it occurs at the extremities.
