JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 29)
A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin$${{\pi t} \over {90}}$$. The ratio of kinetic to potential energy of this particle at t = 210 s will be:
$${1 \over 9}$$
3
2
1
Explanation
K = $${1 \over 2}$$m$${\omega ^2}$$A2cos2$$\omega $$t
U = $${1 \over 2}m{\omega ^2}$$ A2 sin2 $$\omega $$t
$${k \over U}$$ = cot2 $$\omega $$t = cot2 $${\pi \over {90}}$$(210) = $${1 \over 3}$$
Hence ratio is 3 (most appropriate)
U = $${1 \over 2}m{\omega ^2}$$ A2 sin2 $$\omega $$t
$${k \over U}$$ = cot2 $$\omega $$t = cot2 $${\pi \over {90}}$$(210) = $${1 \over 3}$$
Hence ratio is 3 (most appropriate)
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