Answer to Question #229785 in Optics for Mart

Question

Mass m slides without friction inside a hemispherical bowl of radius R. Show that if a particle starts from rest with a small displacement from equilibrium, it moves at simple harmonic motion with an angular frequency equal to that of a simple pendulum of length R. That is ω = √g/R.

Answer 1

As the bowl is frictionless, the particle will execute the simple harmonic motion provided the displacement from the equilibrium to be very small. The problem is categorized under the simple harmonic motion. The simple harmonic wave equation is $a=-ω^2y$ Here, a is the acceleration, y is the displacement and is the angular frequency.

The following figure shows a particle of mass m slides without friction inside a hemispherical bowl of radius R.

The restoring couple on the particle is

F=mgsinθ

The small angular displacement is

$tan θ = \frac{x}{R}$

Here, R is the length of the simple pendulum .

The displacement from the equilibrium position is very small,

So

$tan θ ≈ sin θ$

Substitute $sin θ =\frac{x}{R}$ in the equation and simplify

$F=-\frac{mg}{R}x$

Compare the above equation with the following force equation we get

$F=-kx$

k is the force constant.

Therefore the force constant is

$k = \frac{mg}{R}$

The angular frequency of a particle executing simple harmonic motion is given by

$ω = \sqrt{\frac{k}{m}}$

Substitute k = \frac{mg}{R} in the above equation solve for ω

$ω = \sqrt{ \frac{ \frac{mg}{R} }{m} } \\ = \sqrt{ \frac{g}{R} }$

Proved.