JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 1)
When a particle of mass m is attached to a vertical spring of spring constant k and released, its
motion is described by
y(t) = y0 sin2 $$\omega $$t, where 'y' is measured from the lower end of unstretched spring. Then $$\omega $$ is:
y(t) = y0 sin2 $$\omega $$t, where 'y' is measured from the lower end of unstretched spring. Then $$\omega $$ is:
$$\sqrt {{g \over {{y_0}}}} $$
$${1 \over 2}\sqrt {{g \over {{y_0}}}} $$
$$\sqrt {{{2g} \over {{y_0}}}} $$
$$\sqrt {{g \over {2{y_0}}}} $$
Explanation
_6th_September_Evening_Slot_en_1_1.png)
y(t) = y0 sin2 $$\omega $$t
= $${1 \over 2}{y_0}\left( {2{{\sin }^2}\omega t} \right)$$
= $${1 \over 2}{y_0}\left( {1 - \cos 2\omega t} \right)$$
From comparing standard equation of SHM Amplitude A = $${{{y_0}} \over 2}$$
And frequency = 2$$\omega $$
At equilibrium situation, $${{mg} \over k} = {{{y_0}} \over 2}$$
$$ \Rightarrow $$ $${k \over m} = {{2g} \over {{y_0}}}$$
$$ \therefore $$ 2$$\omega $$ = $$\sqrt {{k \over m}} $$
$$ \Rightarrow $$ 2$$\omega $$ = $$\sqrt {{{2g} \over {{y_0}}}} $$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{2g} \over {4{y_0}}}} $$ = $$\sqrt {{g \over {2{y_0}}}} $$
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