Question 118: Oscillations

Question

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:

Accepted Answer

y(t) = y
0
sin
2
$$\omega $$t
= $${1 \over 2}{y_0}\left( {2{{\sin }^2}\omega t} ight)$$
= $${1 \over 2}{y_0}\left( {1 - \cos 2\omega t} ight)$$
From comparing standard equation of SHM Amplitude A = $${{{y_0}} \over 2}$$
And frequency = 2$$\omega $$
At equilibrium situation, $${{mg} \over k} = {{{y_0}} \over 2}$$
$$ \Rightarrow $$ $${k \over m} = {{2g} \over {{y_0}}}$$
$$ 	herefore $$ 2$$\omega $$ = $$\sqrt {{k \over m}} $$
$$ \Rightarrow $$ 2$$\omega $$ = $$\sqrt {{{2g} \over {{y_0}}}} $$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{2g} \over {4{y_0}}}} $$ = $$\sqrt {{g \over {2{y_0}}}} $$

Answer

$$\sqrt {{g \over {{y_0}}}} $$

Answer

$${1 \over 2}\sqrt {{g \over {{y_0}}}} $$

Answer

$$\sqrt {{{2g} \over {{y_0}}}} $$

Answer

$$\sqrt {{g \over {2{y_0}}}} $$

Oscillations

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:

Choose the correct answer: