Question

A copper rod (length = 2.5 m, radius = 1.60 10-3 m) hangs down from the ceiling. A 8.8-kg object is attached to the lower end of the rod. The rod acts as a "spring", and the object oscillates vertically with a small amplitude. Ignoring the rod's mass, find the frequency f of the simple harmonic motion. Hz

Answer 1

Y = (F / A) (delta L / L₀)

F = (A Y / L₀) delta L

F = k x

k = Y A / L₀

f = [1 / 2 pi] * sqrt (k / m) = [1 / 2 pi] * sqrt (Y A / m L₀)

= [1 / 2 pi] * sqrt (1.12 * 10¹¹ * pi * (1.6 * 10^-3)² / 8.8 * 2.5)

= 32.2 Hz