In my physics notes, it has been given that the damping force increases the period of oscillation. I am unable to understand this part. How is this possible? The only relation I know is that as the damping force increases the amplitude decrease. But how does this contribute to increased time period of oscillation?

  • $\begingroup$ Do you know Fourier transforms? Also, this is very relevant. $\endgroup$ – DanielSank Jan 9 '15 at 8:03
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    $\begingroup$ I am sorry .... I am in 11 th grade and have not yet learnt those stuff .... I can understand none of the equation for in the link provided by u ...😥 I have only been taught the basics... $\endgroup$ – Saravanan Ramesh Jan 9 '15 at 8:18

Suppose we take two identical pendulums. The green one is undamped and the red one is damped:


The force $F_g$ on the green pendulum bob is (approximately) given by the usual simple harmonic oscillator law:

$$ F_g = -kx $$

where $x$ is the displacement, and the acceleration is just $F_g/m$.

Now consider the red pendulum bob. This is damped, so the force on the bob will be the SHO force minus some damping force:

$$ F_r = -(kx - d)$$

where $d$ is some function of the bob velocity and possibly position. Again the acceleration of the red bob is $F_r/m$.

The point is that $F_g > F_r$ (more precisely the magnitude of $F_g$ is greater than the magnitude of $F_r$) and that means the acceleration of the green bob is greater than the acceleration of the red bob. So if we start the two bobs at the same point at time $t = 0$ the red bob will take longer to reach the centre because its acceleration is lower. But if the red bob takes longer to reach the centre than the green bob that must mean its period is longer i.e. its angular frequency is lower.

And that's why damping increases the period of the oscillation.

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