Steady-State Response to Harmonic Forcing

Hello! Welcome to your next lesson on mechanical vibrations.

In the last lesson, we explored how real-world systems dissipate energy through damping, leading to free vibrations that eventually die out. We classified these systems as underdamped, critically damped, or overdamped based on their response to an initial disturbance.

Today, we move from "free" to "forced" vibrations. We'll analyze what happens when a system is subjected to a continuous, oscillating external force. This is a scenario engineers face constantly, from designing buildings to withstand wind or earthquakes, to isolating sensitive equipment from engine vibrations in an aircraft.

Your learning outcome is to determine the steady-state response of a damped system under harmonic forcing. We will find the amplitude and phase of the system's long-term vibration and understand how these are influenced by the system's properties and the forcing frequency.

1. From Free to Forced Vibration

Recall the equation for a free, damped system from our last lesson:
$m\ddot{x}+c\dot{x}+kx=0$
Now, we add an external harmonic force, typically represented as $F(t)=F_0\cos (\omega t)$, where $F_0$ is the force amplitude and $\omega$ is the forcing frequency. The new equation of motion becomes:
$m\ddot{x}+c\dot{x}+kx=F_0\cos (\omega t)$
This is a non-homogeneous second-order differential equation. Just like with the RLC circuits you've studied, its solution has two parts:

1. The Transient Solution ($x_h(t)$): This is the solution to the homogeneous equation ($m\ddot{x}+c\dot{x}+kx=0$). It's the same decaying response we analyzed in the previous lesson. It depends on the initial conditions but dies out over time.
2. The Steady-State Solution ($x_p(t)$): This is the particular solution that depends on the forcing function. After the transient effects have vanished, this is the motion that remains. The system will oscillate at the same frequency as the driving force ($\omega$), but with a different amplitude and a phase lag.

Our goal today is to find this steady-state solution, $x_p(t)$.

2. Solving for the Steady-State Response

Because of your background in electronics engineering, the most efficient and intuitive way to solve for the steady-state response is using complex exponentials, a technique often called the phasor method. This is directly analogous to using phasors and complex impedance to analyze AC circuits.

The video below provides a complete walkthrough of this method. It starts with a quick review of complex numbers (which should be familiar to you) and then systematically derives the steady-state response.

Mechanical Impedance

The video derivation leads to an important concept. When we assume the solution $x(t)=\tilde{X}{e}^{i\omega t}$ and substitute it into the equation of motion, we get:
$(-m{\omega }^2+ic\omega +k)\tilde{X}{e}^{i\omega t}=F_0{e}^{i\omega t}$
$\tilde{X}=\frac{F_0}{k-m{\omega }^2+ic\omega }$
This simplifies a differential equation into an algebraic one. The denominator term is called the mechanical impedance, $Z_m$:
$Z_m=k-m{\omega }^2+ic\omega$
This is the mechanical equivalent of the electrical impedance $Z_e$ in a series RLC circuit:
$Z_e=\frac{1}{i\omega C}+i\omega L+R$
The analogy is direct: force is like voltage, velocity is like current, and mechanical impedance relates them.

For a more formal and concise treatment of this phasor method and the impedance concept, the following document is an excellent reference.

3. Key Formulas: Amplitude and Phase

The final steady-state solution is written as $x_p(t)=X\cos (\omega t-\varphi )$, where $X$ is the amplitude and $\varphi$ is the phase angle by which the displacement lags the force.

By taking the magnitude and angle of the complex amplitude $\tilde{X}$ we found, and non-dimensionalizing the result, we arrive at two of the most important formulas in vibration analysis.

The following image provides a fantastic summary of the system, the key parameters, and the final results.

The key results shown in the image are:

• Magnification Factor (or Amplitude Ratio), $M$: The ratio of the dynamic amplitude $X$ to the static deflection ($F_0/k$) that the same force would cause.
  $M=\frac{X}{F_0/k}=\frac{1}{\sqrt{(1-r^2{)}^2+(2\zeta r{)}^2}}$
• Phase Angle, $\varphi$: The angle by which the displacement lags behind the applied force.
  $\varphi ={\tan }^{-1}\left(\frac{2\zeta r}{1-r^2}\right)$

Where, as before:

• Damping Ratio: $\zeta =\frac{c}{c_{cr}}=\frac{c}{2\sqrt{mk}}$
• Frequency Ratio: $r=\frac{\omega }{{\omega }_n}$ (Ratio of forcing frequency to natural frequency)

4. Interpreting the Response

These formulas tell us everything about the steady-state response. Let's analyze the behavior in different frequency ranges by looking at the curves in the image above.

Let's summarize these regimes:

1. Low-Frequency Region ($r\ll 1$):

   • Amplitude: $M\approx 1$. The amplitude of vibration is approximately the static deflection, $X\approx F_0/k$. The system moves as if the force were applied very slowly.
   • Phase: $\varphi \approx 0^{\circ }$. The displacement is almost perfectly in-phase with the force.

2. High-Frequency Region ($r\gg 1$):

   • Amplitude: $M\to 0$. The amplitude becomes very small. The mass has too much inertia to keep up with the rapidly oscillating force. This is the principle behind vibration isolation. For example, a delicate instrument can be protected from high-frequency "road" vibrations by mounting it on a soft suspension (low $k$, so low ${\omega }_n$ and high $r$).
   • Phase: $\varphi \to 180^{\circ }$. The displacement is completely out-of-phase with the force.

3. Resonance Region ($r\approx 1$):

   • Amplitude: The amplitude $M$ reaches a peak. If the damping $\zeta$ is small, this peak can be extremely large. At exactly $r=1$, $M=1/(2\zeta )$. This is resonance.
   • Phase: $\varphi =90^{\circ }$ at $r=1$. The displacement lags the force by a quarter cycle.

Test your understanding!

An 8 kg avionics package is mounted in an aircraft on a support bracket. The bracket has a stiffness $k=20,000$ N/m and provides a damping ratio $\zeta =0.1$. The aircraft engine creates a harmonic force on the bracket with an amplitude $F_0=40$ N at an operating speed of 2400 RPM.

1. Calculate the natural frequency ${\omega }_n$ of the package-bracket system.
2. Calculate the forcing frequency $\omega$ from the engine RPM.
3. Determine the frequency ratio $r$.
4. Using the formulas, find the amplitude of the steady-state vibration, $X$.

Show answer

1. Natural Frequency:
   ${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{20,000\text{ N/m}}{8\text{ kg}}}=\sqrt{2500}=50\text{ rad/s}$

2. Forcing Frequency:
   $\omega =2400\frac{\text{rev}}{\text{min}}\times \frac{2\pi \text{ rad}}{1\text{ rev}}\times \frac{1\text{ min}}{60\text{ s}}=80\pi \approx 251.3\text{ rad/s}$

3. Frequency Ratio:
   $r=\frac{\omega }{{\omega }_n}=\frac{251.3\text{ rad/s}}{50\text{ rad/s}}\approx 5.03$
   This is in the high-frequency region ($r\gg 1$).

4. Amplitude of Vibration:
   First, find the magnification factor $M$:
   $M=\frac{1}{\sqrt{(1-r^2{)}^2+(2\zeta r{)}^2}}=\frac{1}{\sqrt{(1-5.03^2{)}^2+(2\times 0.1\times 5.03{)}^2}}$
   $M=\frac{1}{\sqrt{(-24.3{)}^2+(1.006{)}^2}}=\frac{1}{\sqrt{590.5+1.01}}\approx \frac{1}{24.32}\approx 0.041$
   Now, find the static deflection ${\delta }_{st}$:
   ${\delta }_{st}=\frac{F_0}{k}=\frac{40\text{ N}}{20,000\text{ N/m}}=0.002\text{ m}=2\text{ mm}$
   Finally, the amplitude is:
   $X=M\times {\delta }_{st}=0.041\times 2\text{ mm}=0.082\text{ mm}$
   The steady-state vibration amplitude is very small (less than a tenth of a millimeter), demonstrating effective vibration isolation.

Conclusion

Today, you've learned how to analyze one of the most fundamental problems in dynamics: the response of a damped system to a continuous harmonic force. We've seen how your background in circuit analysis provides a powerful toolkit for solving these problems.

Key Takeaways:

• The total response of a forced system is the sum of a decaying transient part and a persistent steady-state part.
• The steady-state response occurs at the same frequency as the forcing function, $\omega$.
• The phasor method (using complex exponentials and mechanical impedance) provides an elegant way to find the steady-state amplitude $X$ and phase lag $\varphi$.
• The response is governed by the damping ratio $\zeta$ and the frequency ratio $r=\omega /{\omega }_n$.
• The behavior of the system changes dramatically depending on whether the forcing frequency is much lower than, much higher than, or close to the natural frequency.

Next Lesson Preview:
We've touched upon the most critical aspect of forced vibrations: resonance. When the forcing frequency matches the natural frequency ($r=1$), amplitudes can become dangerously large. In the next lesson, we will focus entirely on resonance, examining its physical significance, its potential for catastrophic failure in structures and machines, and how engineers design to either avoid it or, in some cases, harness it.