# Damped Oscillation and Period

In my school experiment, I wanted to measure the gravitational constant ($$9.81 \ m/s^2$$) by using a pendulum. If we take into account the damped oscillation (i.e. friction forces), does that affect the period?

As far as I know, the position $$x$$ of a particle undergoing Simple Harmonic Motion can be expressed as a function of time: $$x(t) = e^{-t} \sin t.$$

But the answer in the following link says that the friction force might be affected by factors like the velocity, or perhaps the position of the pendulum bob:

Does damping force affect period of oscillation?

In addition, since I am conducting an experiment to find the $$g$$ value, I am concerned if the changing period affects the value of $$g$$.

Could anybody clarify?

• @user47014 But I need the formulae without approximations for some reason. – curious Oct 4 at 12:46
• curious, the amount of damping from air resistance on the pendulum bob will be small enough to ignore. In addition, you should use the equation that alephzero gave you, as your equation for position vs. time is incomplete. – David White Oct 4 at 15:21

The free vibration frequency is affected by damping.

If you assume the damping force is proportional to velocity, the math is well known. See http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html for example, particularly the difference between the damped frequency $$\omega_1$$ and the undamped $$\omega_0$$ in the last equation.

The answer in the link you posted is hand-waving, not rigorous math, but it gives the right physical idea of why the frequency changes.

You can find the amount of damping by measuring how much the amplitude of oscillation decreases from one cycle to the next.

For a pendulum swinging in air, the damping force is not proportional to the velocity (it is more likely to be proportional to the square of the velocity) but for small amounts of damping, you can still fit a curve of the form $$x = A e^{-\gamma t}\cos \omega_1 t$$ to the measured displacements and then find the undamped frequency using $$\omega_0^2 = \omega_1^2 + \gamma^2$$ as in the hyperphysics link.

For a pendulum, the change in frequency is likely to be small, but it is still worth checking if the correction to your experimental data is significant or not.