Simple harmonic motion

Question

A small marble slides along the inside of a cup which has a cross-sectional shape;y=0.100x²,where x is the displacement from the center of the cup and y is the height of the surface above the bottom.if a harmonic oscillation is started,what would be its frequency?

Answer 1

An easy way to work this problem is to note that the energy (which will be independent of time) is given by

E = (1/2) m v² + m g (.1) x²

The time derivative of E is zero so

0 = m v a + m 2 (.1) g x v here a is the acceleration = dv/dt, Also note that the unit of .1 is 1/meter

cancelling out the m and v gives a = - 2 (.1) g x

Letting x = A cos(ω t) and noting that a is the second derivative of x leads to

ω² = 2 (.1) g so ω = sqrt( .2 g )

The linear frequency, f, is ω/( 2 π) so f = sqrt( .2 g ) /(2 π)