A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m. The period of Oscillation of the sphere is (Acceleration due to gravity g = 10 m/s 2 )

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The oscillation of a small sphere in a watch glass of radius of curvature $R$ is equivalent to a simple pendulum of effective length $R$ .

The time period of oscillation is given by $T = 2 π \frac{R}{g}$ .

Given $R = 1.6$ m and $g = 10 m/s^{2}$ .

Substituting the values:

$T = 2 π \frac{1.6}{10}$

$T = 2 π 0.16$

$T = 2 π \times 0.4 = 0.8 π$ s.

Answer: $0.8 π$

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