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?
Harmonic Oscillators – Does Damping Force Affect Period of Oscillation?
frequencyfrictionharmonic-oscillatoroscillators
Related Solutions
For the same reason objects of different masses fall at the same acceleration (neglecting drag): because while the force is proportional to the mass and the acceleration is inversely proportional to mass.
Doing the falling case o avoid having to deal with the vectors in the pendulum we get
$$ a = \frac{F}{m} = \frac{G\frac{Mm}{r^2}}{m} = G\frac{M}{r^2} $$
where $M$ is the mass of the planet, $m$ is the mass of the object you are dropping and $r$ is the radius of the planet.
The mass of the minor object falls out of the kinematics.
The same thing happens in the case of the pendulum: the force includes a factor of $m$, but the acceleration does not.
Mathematical demonstration
It's straightforward to see why this happens if you use a bit of linear response theory. Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says $F(t) = m \ddot{x}(t)$ which gives $$-k x(t) - \mu \dot{x}(t) + F_\text{drive}(t) = m \ddot{x}(t) \, .$$ Dividing through by $m$ and defining $\phi(t) \equiv x(t)/m$, $\omega_0^2 \equiv k/m$, $2 \beta \equiv \mu/m$, and $J(t) \equiv F_\text{drive}(t)/m$, we get $$ \ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2 \phi(t) = J(t) \, .$$ This is a nice general form of the damped driven harmonic oscillator.
Writing $\phi(t)$ as a Fourier transform $$\phi(t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \, \tilde{\phi}(\omega) e^{i \omega t}$$ and plugging into the equation of motion, we find $$\left( - \omega^2 + i 2 \beta \omega + \omega_0^2 \right) = \tilde{J}(\omega)$$ which can be rewritten as $$\tilde{\phi}(\omega) = - \frac{\tilde{J}(\omega)}{\left( \omega^2 - i 2 \beta \omega - \omega_0^2 \right)} \, .$$
Let's take the case where the drive is a cosine, i.e. $J(t) = A \cos(\Omega t)$. In this case $\tilde{J}(\omega) = (1/2)\left(\delta(\omega - \Omega) + \delta(\omega + \Omega) \right)$ so if you work it all out you find $$\phi(t) = \Re \left[ - \frac{A e^{i \Omega t}}{\Omega^2 - i 2 \beta \Omega - \omega_0^2} \right] \, .$$ It's easy to check that $\phi(t)$ has the largest amplitude when $\Omega = \omega_0 \sqrt{1 - 2 (\beta / \omega_0)^2}$, which decreases as $\beta$ increases. Remember that $\beta$ is just proportional to the coefficient of friction $\mu$ so we've shown that more friction makes the peak move to lower frequency.
Resonance
We've shown that the amplitude of the oscillator depends on the damping coefficient. However, this does not mean that the resonance moves to lower frequency. Resonance is a condition defined by unidirectional flow of energy from the drive to the system. It turns out (easy to show with the math we already did) that this happens when $\Omega = \omega_0$, i.e. the drive is at the same frequency as the undamped oscillation frequency. There's already a nice post on this issue which I recommend reading.
Original questions
Why does this happen?
Well, we showed why mathematically. Intuitively it's because the friction takes away kinetic energy so the oscillator doesn't make it as far from equilibrium on each cycle.
And does this imply that at higher damping levels one cannot achieve a higher amplitude by setting the period of the forced oscillation to be equal to the natural frequency?
Assuming constant amplitude of the drive, yes.
This seems strange because, the highest amplitude is achieved when the natural frequency is equal to the driving frequency. But, I guess that the rules are somehow different for the damping case.
Indeed, damping changes things a bit.
Best Answer
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.