It was stated during a physics lecture that a simple harmonic oscillation undergoing light damping would have a period that is slightly greater than it would be without damping but the new period would remain constant throughout. I don't understand it as it seems counter-intuitive that the period would remain constant. 🙁
[Physics] Why is it that period remains constant for an oscillation(SHM) experiencing light damping
harmonic-oscillatoroscillators
Related Solutions
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.
Here $x_m$ represents the displacement of the undamped oscillator; with right initial conditions you can equate that to maximum displacement, i.e, amplitude. Where as the product $x_m e^{-bt/2m}$ is the decaying amplitude of the damped oscillator.
Given initial condition, $x(t=0) = x_0$, you get
$$ x_m = \frac{x_0}{\cos(\phi)} $$
and the initial condition $x'(t=0) = 0$ means the initial velocity of the oscillator is zero, during which the there is no damping takes place, because damping factor $f \propto -\dot{x}$, because of this initial condition you are getting
$$ \frac{b}{2m} = 0 \implies b = 0 $$
the above tells that damping takes place once the oscillator starts moving.
Best Answer
The period stays constant because the solutions can be rescaled at will. So if you have some valid motion $\theta(t)$ describing the pendulum's angle $\theta$ as a function of time $t$, then you can scaled this motion up or down by any factor $s$ and still get a valid motion $s \theta(t)$. This scaling property follows from the niceness (specifically, linearity) of the underlying model. If you remove simplifying assumptions then you lose the scaling property.
Why does the scaling property mean the period is constant? Suppose you have a pendulum and release it from an angle $\theta_1$, then the pendulum will swing away and swing back to a maximum angle $\theta_2$ at some later time $T$. The motion after the time $T$ is the same as if you had simply released the pendulum from the angle $\theta_2$ But now you can use the scaling property to get the future motion of the pendulum. For example, the motion from $T$ to $2T$ must be $\dfrac{\theta_2}{\theta_1} \theta(t-T)$. In fact, this equation must hold for all $t>T$, therefore the pendulum must alway have zero velocity at integer multiples of $T$, and so it shouldn't be hard to believe that the period must be $T$ for all time.