Question 233: Oscillations

Question

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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Step 1: Recall the situation
A small sphere (or particle) oscillates in a spherical surface of large radius $R$. If displaced slightly from equilibrium, the restoring force comes from component of gravity along the surface.
For small oscillations, the motion approximates simple harmonic motion with period:
$ T = 2 \pi \sqrt{\frac{R}{g}} $
where $R$ is the radius of curvature of the spherical surface.
Step 2: Substitute values
Here $R = 1.6 \,	ext{m}$, $g = 10 \,	ext{m/s}^2$.
$ T = 2 \pi \sqrt{\frac{1.6}{10}} $
$ T = 2 \pi \sqrt{0.16} $
$ T = 2 \pi (0.4) = 0.8 \pi \, 	ext{s}. $
Step 3: Final Answer
So the correct option is:
Option D: $0.8 \pi \, \mathrm{s}$
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Answer

$0.2 \pi \mathrm{~s}$

Answer

$0.4 \pi \mathrm{~s}$

Answer

$0.6 \pi \mathrm{~s}$

Answer

$0.8 \pi \mathrm{~s}$

Oscillations

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 \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: