Kansas State University

Discussion

An LRC circuit with an AC power source includes the following components. As current flows through the loops of a coil, the spinning current induces a magnetic field. If the current flow changes, then the magnetic field changes. A changing magnetic field inside the loops will then induce a current. This acts like mass in a spring-mass system, since the induced current opposes the change in the current, just as mass measures opposition to change in momentum. A coil has "inductance" $L$, which is measured in henrys. Voltage drop across a coil is given by $\displaystyle V=L\frac{dI}{dt}=L\frac{d^2Q}{dt^2}$.

A resistor does what it sounds like, it resists the flow of current. Resistance $R$, measured in ohms, plays the role of friction. The voltage drop across a resistor is $\displaystyle V=RI=R\frac{dQ}{dt}$.

A capacitor stores charge on parallel plates separated by a dielectric. The amount of charge is proportional to the impressed voltage, $Q=CV$, or $\displaystyle V=\frac{1}{C}Q$. Capacitance $C$ is measured in farads, which are coulombs per volt. Since the charge will be released as the voltage drops, the capacitor plays the role of the spring, storing energy and then releasing it later. In fact, $1/C$ is called the "elastance" of the circuit and plays the role of the spring constant (and has units of "daraf", though reciprocal farad is the more boring preferred term).

Finally, the AC power source produces an alternating voltage described by a sinusoidal function, $E(t)=V_0\cos(\omega t)$. $V_0$ is the peak amplitude and $\omega$ is the circular frequency. Note that when voltages are given, you typically use the root-mean-square amplitude rather than the peak amplitude, so a 120 volt supply would correspond to a peak amplitude of $120\sqrt2\approx 170$ volts. Similarly, frequency is usually given in Hertz, which are cycles per second, while circular frequency is measured in radians per second, so a 60 Hertz supply has $\omega=60\times2\pi=120pi$ since 2π radians equals one full cycle.

We now consider either an LRC circuit or a spring-mass system subject to an external force. These are mathematically identical if you just match things up as discussed above. And just as an AC power source supplies voltage in the form $E(t)=V_0\cos(\omega t)$, the most interesting forces are periodic forces, which we will assume take the form $F(t)=F_0 \cos(\omega t)$ where $F$ is force, $t$ is time, $F_0$ is amplitude and $\omega$ is the circular frequency. If you are looking at how vibrations from a motor running or any other periodic phenomenon affects a structure, you will be led to these sorts of forces. In what follows, we will use the labels for a spring-mass system, since that is more familiar to most students. Just remember everything works the same for circuits.

We start with the undamped case $(c=0)$. We have solved the homogeneous problem in the previous section, obtaining $A\cos(\omega_0(t + \phi))$ where $\omega_0 =\sqrt{k/m}$ is called the natural frequency of the spring-mass system. We will now consider the particular solution of $$m\frac{d^2x}{dt^2}+kx=F_0\cos(\omega t)$$ Since $F_0 \cos(\omega t) = \Re[F_0 e^{i\omega t}]$, we will try to find the particular solution for the complex problem and then take the real part. So we guess $$ \begin{align} z&=ae^{i\omega t} \\ z'&=i\omega ae^{i\omega t} \\ z''&=-\omega^2ae^{i\omega t} \end{align} $$ So we get $$ \begin{align} -m\omega^2ae^{i\omega t}+kae^{i\omega t}&=F_0e^{i\omega t} \\ a&=\frac{F_0}{-m\omega^2+k}=\frac{F_0}{m(\omega_0^2-\omega^2)} \end{align} $$ and taking the real part of $ae^{i\omega t}$ yields a particular solution of $$ x(t)=\frac{F_0}{m(\omega_0^2-\omega^2)}\cos(\omega t)$$ Note that the particular solution has the same frequency as the forcing function but that the amplitude is divided by a factor $m(\omega_0^2 - \omega^2)$. We would expect to see that mass would be inversely proportional to the displacement generated by a fixed force, but it may be something of a surprise that the frequency plays such a large role. The underlying physical intuition is that the spring wants to oscillate at its natural frequency, and the closer the forcing function is to the natural frequency, the more the forcing function and the spring will work together and not in opposition, hence the larger amplitude of the response. The general solution is then $$x(t)=\frac{F_0}{m(\omega_0^2-\omega^2)}\cos(\omega t)+A\cos(\omega_0(t+\phi))$$ If we plot out the solution curves, we will see the phenomenon of beats arising. This is most easily seen if $\omega$ and $\omega_0$ are close. Then the amplitude of the solution will steadily increase as the forcing function pours ever greater amounts of energy into the system. However, if the forcing function is not exactly in sync with the natural frequency of the system (that is $\omega\ne\omega_0$) then eventually the forcing function will become out of phase with the natural frequency. Then the force applied to the system will reduce the amplitude and the ``beat'' will die off. At the end of the beat, the forcing function will have worked its way back in sync with the natural frequency and the pattern will start all over again.

Of course, if $\omega=\omega_0$ , then we have divided by zero and our solution is not valid for this case. Here, we get a particular solution of the form $$x(t)=\frac{F_0}{2m\omega_0}t\sin(\omega_0t)$$ You may check that this is indeed a particular solution. Note that in this case, the amplitude grows linearly with $t$. Since the forcing function and the natural frequency are exactly in sync, the forcing function always acts to add energy to the system and the amplitude grows without bound, at least mathematically. In real life, the system will tear itself apart eventually as the amplitude gets too large. This is the phenomenon of resonance.

Next we consider what happens when we include damping $(c>0)$. We will consider just the underdamped case (the overdamped is similar). The homogeneous solution is $$ x(t)=Ae^{-(c/2m)t}\cos(\omega_1(t+\phi)) $$ where $\omega_1 = \sqrt{\omega_0^2-(c/2m)^2}$ and $\omega_0=\sqrt{k/m}$ is the natural frequency. Note that if $c$ is small then the spring-mass system oscillates with circular frequency $\omega_1$ very close to the natural frequency $\omega_0$.

Now when we find the particular solution, we again use $F_0 \cos(\omega t)=\Re[F_0 e^{i\omega t}]$ and solve for a particular solution to the complex problem and take the real part. We guess $$\begin{align} z &= ae^{i\omega t} \\ z' &= i\omega ae^{i\omega t} \\ z'' &= -\omega^2 ae^{i\omega t} \end{align} $$ and so we find $$ \begin{align} -m\omega^2ae^{i\omega t}+ci\omega ae^{i\omega t}+kae^{i\omega t} &=F_0e^{i\omega t} \\ \left((k-m\omega^2)+ic\omega\right)ae^{i\omega t} &=F_0e^{i\omega t} \\ a&=\frac{F_0}{(k-m\omega^2)+ic\omega}\\ a&=\frac{F_0}{m(\omega_0^2-\omega^2)+ic\omega} \end{align} $$ Here there is no danger of dividing by zero. Even if $\omega_0=\omega$, $c>0$ so the denominator is non--zero. Now let $R = \sqrt{m^2(\omega_0^2-\omega^2)^2+c^2\omega^2}$ be the modulus of $m(\omega_0^2 - \omega^2 ) + ic\omega$ and let $\Phi = \arctan(c\omega/[m(\omega_0^2 - \omega^2 )])$ be the argument of $m(\omega_0^2-\omega^2)+ic\omega$. Then the real particular solution will be $$\Re\left[\frac{F_0}{Re^{i\Phi}}e^{i\omega t}\right] =\Re\left[\frac{F_0}{R}e^{i(\omega t-\Phi)}\right] =\frac{F_0}{R}\cos(\omega t-\Phi)$$ The general solution is $$\frac{F_0}{R}\cos(\omega t-\Phi)+e^{-ct/2m}A\cos(\omega_1t+\phi)$$

The first term (the particular solution) is called the steady state solution and the second term (the homogeneous solution) is called the transient solution. Note that the amplitude of the transient term decays exponentially, so it will become unimportant as $t$ grows large, hence the name transient. The steady state term has a constant amplitude, hence the name steady state. With positive damping, $R$ will never be zero so the solution will not have amplitude that increases without limit. On the other hand, if $\omega_0=\omega$ then $R=c$ and if $c$ is small, it is still possible for the response to be sufficiently large to break the spring-mass system.