A particle undergoing simple harmonic motion has a dependent displacement given by x(t) = $x(t) = A\sin \frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at t = 210s will be:Option 1)  Option 2)$\frac{1}{3}$  Option 3)$\frac{1}{9}$Option 4)  1

Kinetic energy in S.H.M. -

$K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}$

- wherein

$K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}$

Potential energy in S.H.M. -

$P.E.= \frac{1}{2}Kx^{2}$

- wherein

Where $K= m\omega ^{2}$

Option 1)

Option 2)

$\frac{1}{3}$

Option 3)

$\frac{1}{9}$

Option 4)

1

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