A bead on a smooth vertical ring of radius 0.5 m rotating about the vertical diameter sits in equilibrium at θ = 6 0 ∘ from the downward vertical. Find the period of small oscillations about this posi

Question

A bead on a smooth vertical ring of radius 0.5 m rotating about the vertical diameter sits in equilibrium at $θ = 6 0^{\circ}$ from the downward vertical. Find the period of small oscillations about this position, in s ( $g = 9.8$ ).

Answer:

1.16

Solution:

$T = 2 π \frac{r cos θ}{g sin ^{2} θ} = 2 π \frac{0.25}{9.8 ( 0.75 )} = 1.159$ s.

TestHub · Accepted Answer

T = 2 π g sin 2 θ r cos θ ​ ​ = 2 π 9.8 ( 0.75 ) 0.25 ​ ​ = 1.159 s.

Physics - SHM Question with Solution | TestHub

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