Problem — Lead and cork pendulums

Problem

Two pendulums identical in length, but one has a lead sphere and the other a cork sphere of the same size. Which oscillates with the greater period? Why does the result change if the amplitude is not small? Discuss what role air friction plays.

Solution

Ideal case (small oscillations, no friction). The period of the simple pendulum is:

$T = 2\pi\sqrt{\frac{\ell}{g}}$

The mass does not appear: the inertial mass that resists and the gravitational mass that pulls cancel out. Hence, for equal length, the two pendulums have the same period:

$\ev{T_\text{piombo} = T_\text{sughero}}$

Role of air friction. Air resistance depends on the shape and size of the sphere, not on its mass. For equal area, the same braking force acts on a small mass (cork) and on a large mass (lead). On the cork the relative effect is much stronger: the cork loses amplitude quickly and stops sooner. If the damping is strong, the effective period of the cork increases slightly relative to the ideal value.

Non-small amplitude. For large amplitudes the period increases relative to the small-oscillation value, with the approximate correction:

$T \approx T_0 \left(1 + \frac{\theta_0^2}{16}\right)$

This effect holds for both. However, the cork, damping quickly, soon reduces its own amplitude: its period therefore tends to return rapidly towards the small-oscillation value, while the lead retains large amplitudes for longer.

Links

Topics: Oscillations and harmonic motion Concepts: Simple pendulum Skills: Limiting-case and thought-experiment analysis Objects: Simple pendulum