Question

A small drop of mercury oscillates simple harmonically inside a watch glass whose radius of curvature is 2.5 m. Find the period of the motion. [g = 9.8m/s²]

Answer 1

Data : R = 2.5 m, g = 9.8 m/s² 

Consider a small drop of mercury on a watch glass of radius of curvature R.

Away from its equilibrium position O, its weight m\(\vec g\) is resolved into two perpendicular components : mg cos θ normal to the concave surface and mg sin θ tangential to the surface, mg cos θ is balanced by the normal reaction (\(\vec N\)) of the surface while mg sin θ constitutes the restoring force that brings the drop back to O. If θ is small and in radian,

restoring force, F = ma = – mg sin θ

= – mg θ

= -mg \(\frac{x}R\)

∴ The acceleration per unit displacement, \(|\frac{a}x|\) = \(\frac{g}R\)

∴ The period of the motion, T = \(\frac{2\pi}{\sqrt{|a/x}|}\) = 2\(\pi\) \(\sqrt{\frac{R}g}\)

Data : R = 2.5 m, g = 9.8 m/s² 

∴ The period of oscillation is

T = 2 × 3.142 \(\sqrt{\frac{2.5}{9.8}}\) = 6.284 × 0.5051 = 3.174s.