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

Question

The root mean square velocity ($v_{\text{rms}}$) of a gas is given by:

Answer 1

Step 1: Restoring force:
For a small angle $\theta$, the restoring force is $mg\sin\theta\approx mg\dfrac xR$.

Step 2: Angular frequency:
$\omega=\sqrt{g/R}=\sqrt{10/1.6}=2.5$ rad/s.

Step 3: Period:
$T=\dfrac{2\pi}{2.5}=0.8\pi$ s. Option (C).

Final Answer:
omega = 2.5 rad/s gives T = 0.8 pi. \[ \boxed{C} \]

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 \(g = 10\text{ m/s}^2\))

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on simple harmonic motion

Questions Asked in MHT CET exam