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

The formula for the time period is:

$T = 2 π \frac{R}{g}$

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

Substituting the values:

$T = 2 π \frac{1.6}{10}$

$T = 2 π 0.16$

$T = 2 π \times 0.4$

$T = 0.8 π s$

Answer: $0.8 π s$

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