Forced Harmonic Vibration and Resonant Frequency 📂Classical Mechanics

Forced Harmonic Vibration and Resonant Frequency

Forced Harmonic Oscillation¹

Harmonic motion, such as the movement of an object hanging from a spring, occurs when no external forces other than the spring constant $k$ ’s restoring force are acting. This is called simple harmonic oscillation when excluding other external forces like air resistance or any friction. If there is an external force proportional to the velocity, like friction, it’s referred to as damped harmonic oscillation. When an external periodic force, driving force, is applied, it is known as forced harmonic oscillation, and its equation of motion is as follows.

$\begin{equation} m\ddot{x}= -kx -c\dot{x} +F_{0}\cos \omega t \end{equation}$

$-c\dot{x}$ : Damping term, an external force proportional to velocity (friction, etc.)
$F_{0}\cos \omega t$ : Periodic external force

In a forced oscillation system, a peculiar phenomenon called resonance occurs. Regardless of the amplitude $F_{0}$ of the driving force, and even when fixed, when the driving frequency $\omega$ is close to the system’s natural frequency $\omega_{0}$ , the amplitude increases. A good example to understand this easily is a swing. When swinging alone, the number of swings per second is the system’s natural frequency. If someone pushes from behind, their pushing force acts as the driving force, and the number of pushes per second is the driving frequency. As commonly experienced, pushing the swing when it’s furthest back to start moving forward will make it go the furthest. Mechanically, this can be explained as ’the phenomenon where the amplitude increases as the driving frequency gets closer to the natural frequency is called resonance.’ Let’s now look at the case of forced harmonic oscillation with and without damping.

Without Damping

Without the damping term, $(1)$ is as follows.

$\begin{equation} m\ddot{x} +kx =F_{0}\cos \omega t \label{eq2} \end{equation}$

The solution to this differential equation is represented by $\cos$ . Or, simply thinking ’to make a function plus its twice differentiated form equal cosine, that function must be cosine’ is fine. Therefore, let’s assume the solution is as follows.

$x(t)=A \cos (\omega t -\phi)$

By substituting this into $(2)$ , we get:

$\begin{align*} &&- m\omega^{2}A \cos (\omega t -\phi)+kA\cos (\omega t - \phi)&=F_{0}\cos \omega t \\ \implies && (k- m\omega^{2})A \cos (\omega t -\phi)&=F_{0}\cos \omega t \end{align*}$

For this equation to hold, $\phi=0$ and $\phi=\pi$ must be true. This is because only these two cases can make the same shape by including changing the sign of the cosine function values when translated parallel. Then, the amplitude $A$ is as follows.

$A= \begin{cases} \dfrac{F_{0}/m}{{\omega_{0}}^{2}-\omega^{2}} & \phi=0,&\omega < \omega_{0} \\[1em] \dfrac{F_{0}/m}{\omega^{2}-{\omega_{0}}^{2}} & \phi=\pi,&\omega > \omega_{0} \end{cases}$

Here, $\omega_{0}=\sqrt{k/m}$ is the system’s natural frequency. By setting $\omega_{0}=4$ , $F_{0}/m = 1$ , and drawing a graph of the amplitude according to the driving frequency, it looks like below.

With Damping

Including the term differentiated once, i.e., velocity in the equation makes it preferable to represent the driving force as an exponential function. If the vibration has amplitude $F_{0}$ and frequency $\omega$ , it’s represented by $F_{0}e^{i\omega t}$ , so $1$ is as follows.

$\begin{equation} m\ddot{x}+c\dot{x}+kx = F_{0}e^{i\omega t} \label{eq3} \end{equation}$

Then, similarly to the undamped case, let’s assume the solution is as follows.

$x(t)=A e^{i(\omega t-\phi)}$

By substituting this into $\eqref{eq3}$ , we obtain the following equation.

$-m\omega ^{2}Ae^{i(\omega t -\phi)}+i\omega cAe^{i(\omega t -\phi)}+kAe^{i(\omega t - \phi)}=F_{0}e^{i \omega t}$

Multiplying both sides by $e^{-i (\omega t -\phi)}$ yields:

$-m\omega ^{2}A+i\omega cA+kA=F_{0}e^{i \phi}=F_{0}(\cos \phi +i \sin \phi)$

The right side is valid due to Euler’s formula. Dividing this equation into real and imaginary parts gives:

$\begin{equation} \begin{aligned} A(k-m\omega^{2}) = F_{0}\cos \phi \\ c\omega A = F_{0} \sin \phi \end{aligned} \label{eq4} \end{equation}$

Dividing the lower equation by the upper one yields the following condition for the phase difference $\phi$ :

$\frac{c\omega}{k-m\omega ^{2}}=\tan \phi$

Moreover, from $\eqref{eq4}$ and using that $\sin ^{2} \phi + \cos ^{2} \phi=1$ , we get the following equation.

$A^{2}(k-m \omega ^{2})^{2} + c^{2} \omega^{2} A^{2} =F_{0}^{2}$ Organizing this for the amplitude gives:

$\begin{align*} A(\omega) &= \frac{F_{0}}{\sqrt{(k-m \omega ^{2})^{2} + c^{2} \omega^{2}}} \\ &= \frac{F_{0}/m}{\sqrt{(k/m- \omega ^{2})^{2} + c^{2} \omega^{2}/m^{2}}} \\ &= \frac{F_{0}/m}{\sqrt{({\omega_{0}}^{2}- \omega ^{2})^{2} + 4\gamma ^{2}\omega^{2}}} \end{align*}$

Here, $\omega _{0}$ is the system’s natural frequency, and $\gamma=\frac{c}{2m}$ is the damping factor. Now, let’s find the frequency at which resonance occurs. When the denominator is $0$ , $A(\omega)$ diverges, so let’s set it as follows.

$({\omega_{0}}^{2}-\omega^{2})^{2}+4\gamma ^{2} \omega^{2} = 0$

$0$ remains the same even when differentiated, so differentiating the above equation with respect to $\omega$ gives:

$2({\omega_{0}}^{2}-\omega^{2})(-2\omega) + 8\gamma^{2}\omega=0$

Organizing it for $\omega$ yields:

${\omega_{r}}^{2}=\omega^{2}={\omega_{0}}^{2}-2\gamma^{2}$

When $\omega$ is as above, resonance occurs because $A(\omega)$ diverges, so this frequency is called the resonant frequency and denoted as $\omega_{r}$ . $r$ is derived from resonance. The smaller the damping factor $\gamma$ is, the closer the resonant frequency is to the natural frequency $\omega_{0}$ . When $\omega_{0} = 2$ , the graph of amplitude depending on the value of $\gamma$ looks as follows. The smaller $\gamma$ gets, the closer the peak of the graph moves to $2$ .