Why does adding a term $5f'(t)$ to $5f''(t)+10f(t)=0$ cause damping?

Question

So we have a differential equation to model an oscillator: $$5f''(t)+10f(t)=0$$

Where the initial conditions are $f(0)=0$ and $f'(0)=4$.

It is given that $f(t) = \frac{2\sqrt 2}{5}\sin\sqrt2 t$.

Now, if we make make a slight change to the original equation and add in a multiple of $f'(t)$, specifically: $$5f''(t)+5f'(t)+10f(t)=0$$

The initial conditions are still same as before, but it appears adding this $f'(t)$ introduces damping. I solved the equation above to yield: $$f(t)=\frac{8e^{-\frac t2}}{\sqrt 7}\sin \frac{\sqrt 7}{2}t$$

My question is this: why does the addition of a multiple of $f'(t)$ create this effect?

Answer 1

If $f(t)$ stands for the displacement of the oscillator from its equilibrium, then the first derivative $f'(t)$ will be the velocity and the second derivative $f''(t)$ the acceleration (which, according to Newton's $2^{nd}$ law of motion, is always proportional to the resultant force exerted on the oscillator).

So, your first equation $5f''(t)+10f(t)=0$ says that: $\sum F(t)=-kx(t)$, ($k$:constant) which is Hooke's law i.e. the necessary and sufficient condition for a free (undamped) oscillation.

On the other hand, your second equation $5f''(t)+5f'(t)+10f(t)=0$ says that: $\sum F=-kx(t)-bu(t)$, ($k,b$:constant) i.e. there is a component of the net force proportional to the velocity: $-bu(t)$, which is exactly the damping force. (Usually frictional forces, air resistance and more generally various resistances to motion are modelled by summands proportional to the velocity or to some power of it).

Answer 2

Let's write the differential equation like this: $$f''(x)=-2f(x)-f'(x).$$ The acceleration of the oscillator is equal to $-2$ times its position minus its velocity. According to the initial conditions, this means that the initial velocity is positive, position is zero, so the initial acceleration is negative. Since velocity and acceleration have opposite sign, the oscillator is slowing down. In the original system, the acceleration at the beginning was zero, so when we add the damping term, the oscillator slows down more abruptly than before. This means that the first maximum of position is a little less with damping than otherwise. Now the oscillator enters the speed up phase. But since the Oscillator's maximum is not as extreme as before, its acceleration is smaller in this phase, so when it reaches its maximum velocity, its speed is also smaller. The oscillator is losing energy. The pattern continues forever. The oscillator loses energy and decreases in amplitude exponentially.

Answer 3

The real part of the root leads to exponentially growing or decaying behavior depending on the sign. The imaginary part leads to oscillatory behavior. In your case, complex roots with a negative real part lead to a damped oscillator.

With $f'' = -cf$, one could expect a purely imaginary root (if $c>0$). That's a perfect oscillator, which makes sense. The acceleration is dependent on past position. Think about a mass on a spring whose force $F=-kx$. It's acceleration at any given time depends on its position.

With $f'' = -bf'-cf$, we could expect real or complex roots, depending on $b$. If the root is complex, then we will have an oscillator, but it's being modified by the function's dependence on velocity. If the real part is negative, think mass on a spring, but the spring is now subject to some frictional or fluid resistance. Air resistance is sometimes modeled as proportionate to the velocity, $F=-\mu v$. This would damp your oscillator.

Answer 4

The straightforward answer is that this part add extra damping effect of whole system.

While in proof, we'd better solve the equation explicitly:

For these kinds of homogeneous ODE with constant factors, we could manage to re-write the equation into: $$p(\partial_t)f(t)=0$$ $p(\partial_t)$ is a polynomial function of operator $\partial_t$.

Using the $\lambda_i$ root notation:$$p(\partial_t)=\prod_{i}(\partial_t -\lambda_i)$$ keep in mind that constant commute with derivation operator $\partial_t$, just do integrals in product order. $$p(\partial_t)f=(\partial_t-\lambda_1)\prod_{i=2}^{n}(\partial_t -\lambda_i)f$$

There is a trick that: $$(\partial_t-\lambda)f=e^{\lambda t}\partial_{t}(e^{-\lambda t}f)$$ After integral from initial situation the $e^{\lambda t}$ part will remain. Thus the spectrum of $\lambda_i$ decides the solve's form as $e^{\lambda t}$ times polynomials(check by hands) : $$\sum _{i}c_{i}(t)e^{\lambda_i t}\ \ \ \ Rank(c_{i}(t))=Rank(\lambda_i)$$.

In first equation the $p_{0}(\partial_t)=5\partial_t ^{2}+10 $ has pure imaginary roots, while after adding the $5\partial_t$ part we could get two complex roots $-\frac{1}{2}\pm i\frac{\sqrt{7}}{1}$, the possible real solution will definitely contain damping factor $e^{-\frac{t}{2}}$.

Answer 5

This happens due to damping dynamics of second order systems. The coefficient when zero gives undamped harmonic motion. When non-zero, there are 3 types of damping ( over damped, critically damped under damped).The graphs are very instructive.

2ndOrderOscillations