A circular wire frame of radius R is rotating about its fixed vertical diameter. A bead on the wire remains at rest relative to the wire at a position in which the radius makes an angle $θ$ with the vertical (see figure). There is no friction between the bead and the wire frame. Prove that the bead will perform SHM (in the reference frame of the wire) if it is displaced a little from its equilibrium position. Calculate the time period of oscillation.

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$2 π \sqrt{\frac{R S i n θ}{g}}$

$2 π \sqrt{\frac{Rcos θ}{g}}$

$π \sqrt{\frac{Rcos θ}{g}}$

$2 π \sqrt{\frac{2 Rcos θ}{g}}$

answer is D.

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Detailed Solution

In reference frame of the wire, the equilibrium of bead gives
${mω}^{2} rcos θ = mgsin θ$ [Equilibrium along tangent]
$\Rightarrow ω^{2} r = gtan θ . . . . . (i)$
If $θ$ is increased by a small amount, say $∆ θ$ , the tangential force [along upward tangent] also changes.

$F_{t} = {mω}^{2} rcos θ - mgsin θ$
${ΔF}_{t} = \frac{d}{dθ} [{mω}^{2} rcos θ - mgsin θ] Δθ = [- {mω}^{2} rsin θ - mgcos θ] Δθ$
$\begin{array}{l} ∴ m \frac{d^{2} (RΔθ)}{{dt}^{2}} = [- {mω}^{2} rsin θ - mgcos θ] Δθ \\ ∴ \frac{d^{2} (Δθ)}{{dt}^{2}} = \frac{1}{R} [- gtan θ \cdot \sin θ - gcos θ] Δθ = - \frac{g}{R} [\frac{\sin^{2} θ}{\cos θ} + \cos θ] Δθ = - (\frac{g}{Rcos θ}) Δθ \end{array}$
Hence motion is SHM.
$ω^{2} = \frac{g}{Rcos θ} \Rightarrow T = 2 π \sqrt{\frac{Rcos θ}{g}}$

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