Math 555: Differential Equations

3.8 Forced Periodic Vibrations

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3.8.1 Forced Periodic Vibrations

Consider Nonhomogeneous the 2nd-Order Differential Equations of the form

$$y'' + p(t)y' + q(t)y = g(t)$$
Under normal situations our resistance term $p(t) > 0$. There are negative differential resistance components used in electronic oscillators and amplifiers. However these are nonlinear devices that exhibit negative differential resistance over a limited portion of their voltage or current range. In this section we limit ourselves to periodic or quasi-periodic motion. That means that $p(t) \ge 0$ and the discriminant $p(t)^2 - 4q(t) \lt 0$. This gives us the characteristic equation

$$d^2 + p(t)d + q(t) = 0$$
with roots

$$r = \dfrac{-p(t)\pm\sqrt{p(t)^2 - 4q(t)}}{2}.$$
While we will need to study variable coefficients in the future in this chapter we consider only the case where our differential equation has constant coefficients.

Recall that if we have constant coefficients

$$ ay'' + by' + cy = g(t)$$
then the roots of the characteristic polynomial are given by

$$r = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = -\dfrac{b}{2a} \pm\sqrt{\dfrac{b^2-4ac}{4a^2}}.$$
If we define $\lambda = \dfrac{b}{2a} \geq 0$ then

$$r = -\lambda\pm\sqrt{\lambda^2 - \dfrac{c}{a}},$$
If we have oscillatory motion, then $\lambda^2 - \frac{c}{a} < 0$. Defining the frequency response as $\mu = \sqrt{\frac{c}{a}-\lambda^2}$ gives us

$$r = -\lambda\pm i\mu.$$
This results in the homogeneous solution

$$y(t) = e^{-\lambda t}\Big(c_1\cos\left(\mu t\right) + c_2\sin\left(\mu t\right)\Big),$$
If $\lambda=0$, that is if the differential equation is undamped, then the solution defines steady state periodic motion. In this case the response frequency $\mu = \sqrt{\frac{c}{a}}$ is called the natural frequency and denoted

$$\omega_0 = \sqrt{\frac{c}{a}}.$$

Steady State, Undamped Periodic Vibration

If $\lambda>0$, that is if the differential equation is damped, then the motion will be defined by a quasi-periodic transient solution which will decrease to zero given no other force acting on the system.

Transient, Damped Periodic Vibration

For this reason we call our differential equation forced if the right-hand side $g(t)\neq 0$ and we call the function $g(t)$ a forcing function . Adding a forcing function can result in steady state solutions, transient solutions or both! We call the forcing function the input or signal , and we call the steady state solution the forced response or simply response to the forcing function.

We can define a function whose input is the signal $g(t)$, and its output the response, that is the solution to the inhomogeneous (another term for nonhomogeneous) differential equation. This is not a very good function because for each input $g(t)$ we get infinitely many outputs, the family of solutions to the differential equation. We can define a better function for an initial value problem that satisfies the hypothesis of Theorem 3.2.1 . This function maps each input $g(t)$ to the response or unique solution to the initial value problem $\phi(t)$.

If the input is periodic and the frequency of the input is too close to the natural frequency of the differential equation then the output exhibits resonance . Resonance can be either good or bad. The design of a bridge or a building should avoid resonance, but an antenna or detector would use resonance to amplify small input signals.

3.8.2 Determining Steady State Response

To find the steady state response , we need only find the particular solution for a differential equation. This still requires us to solve the homogeneous equation, because we need to ensure that our guess for undetermined coefficients is not linearly dependent on the homogeneous solution.

Example 3.8.1

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant of $k = \frac{1}{4}$ lb$\cdot$s/ft and is acted on by an external force of $4\cos(2t)$ lb. Determine the steady state response of this system.

Solution

Our mathematical model is given by

$$mu'' + \gamma u' + ku = F_0\cos(\omega t).$$
We need to find the values of our constants $m$, $\gamma$, and $k$. Using the approximation $g = 32$ ft/s$^2$, we have for this system

$$\begin{align*} m &= \dfrac{8\,\text{lb}}{32\,\text{ft/s$^2$}} = \dfrac{1}{4}\,\text{lb$\cdot$s$^2$/ft}\\ \\ \gamma &= \dfrac{1}{4}\,\text{lb$\cdot$s/ft} \\ \\ k &= \dfrac{8\,\text{lb}}{\frac{1}{2}\,\text{ft}} = 16\,\text{lb/ft} \\ \\ F_0\cos(\omega t) &= 4\cos(2t) \end{align*}$$
Plugging these values into our mathematical model yields

$$ \dfrac{1}{4}u'' + \dfrac{1}{4}u' + 16u = 4\cos(2t), $$
then multiplying by $4$ to simplify gives

$$ u'' + u' + 64u = 16\cos(2t). $$
Since the characteristic equation for the homogeneous case is $r^2 + r + 64 = 0$, we see that the roots are

$$r = -\frac{1}{2} \pm i\frac{\sqrt{255}}{2}.$$
Clearly, the forcing function $4\cos(2t)$ is not linearly dependent on this, so we may proceed to applying the method of undetermined coefficients:

$$\begin{align*} u_p(t) &= A\cos(2t) + B\sin(2t) \\ \\ u_p'(t) &= 2B\cos(2t) - 2A\sin(2t) \\ \\ u_p''(t) &= -4A\cos(2t) - 4B\sin(2t) \\ \\ 64u_p(t) &= 64A\cos(2t) + 64B\sin(2t) \\ \\ 16\cos(2t) &= \left(60A + 2B\right)\cos(2t) + \left(-2A + 60B\right)\sin(2t) \end{align*}$$
Now, we may use this to construct a matrix equation and apply Cramer's rule to find the coefficients $A$ and $B$

$$\begin{align*} \left[\begin{array}{cc} 60 & 2 \\ -2 & 60 \end{array}\right]\left[\begin{array}{c} A \\ B \end{array}\right] &= \left[\begin{array}{c} 16 \\ 0 \end{array}\right] \\ \\ \left|\begin{array}{cc} 60 & 2 \\ -2 & 60 \end{array}\right| &= 3600 + 4 = 3604 \\ \\ A = \dfrac{\left|\begin{array}{cc} 16 & 2 \\ 0 & 60 \end{array}\right|}{3604} &= \dfrac{960}{3604} = \dfrac{240}{901} \\ \\ B = \dfrac{\left|\begin{array}{cc} 60 & 16 \\ -2 & 0 \end{array}\right|}{3604} &= \dfrac{32}{3604} = \dfrac{8}{901} \end{align*}$$
Our particular solution, and hence steady state response, is

$$ u_p(t) = \dfrac{240}{901}\cos(2t) + \dfrac{8}{901}\sin(2t).$$

3.8.3 Maximizing the Response's Amplitude

Example 3.8.2

Using the mathematical model from Example 3.8.1 , determine the mass $m$ for which the amplitude of the steady state response is maximum.

Solution

We are still using the 2nd-order nonhomogeneous linear differential equation

$$mu'' + \gamma u' + ku = F_0\cos(\omega t),$$
and still have that

$$\begin{align*} \gamma &= \dfrac{1}{4}\,\text{lb$\cdot$s/ft} \\ \\ k &= \dfrac{8\,\text{lb}}{\frac{1}{2}\,\text{ft}} = 16\,\text{lb/ft} \\ \\ F_0\cos(\omega t) &= 4\cos(2t). \end{align*}$$
However, our mass is now in slugs $m$. This gives us the equation

$$mu'' + \dfrac{1}{4}u' + 16u = 4\cos(2t).$$
In this case we need to consider the steady state amplitude $R = R(m)$ as a function of the mass $m$. Then we will need to differentiate the function $R(m)$ and set the derivative equal to zero to find a critical point. We will need the second derivative to be negative for us to have a maximum value of $R$ and the critical mass $m$. It is easier to make our computations with symbols and plug in our constants at the end!

Starting with our mathematical model,

$$ mu'' + \gamma u' + ku = F_0\cos(\omega t), $$
our guess for the particular solution and its derivatives are

$$\begin{align*} u_p(t) &= A\cos(\omega t) + B\sin(\omega t) \\ \\ u_p'(t) &= B\omega\cos(\omega t) - A\omega\sin(\omega t) \\ \\ u_p''(t) &= -A\omega^2\cos(\omega t) - B\omega^2\sin(\omega t). \end{align*}$$
We then multiply each element of the particular solution by the appropriate constant for our ODE, plug everything in, and collect terms:

$$\begin{align*} ku_p(t) &= kA\cos(\omega t) + kB\sin(\omega t) \\ \\ \gamma u_p'(t) &= \gamma\omega B\cos(\omega t) - \gamma\omega A\sin(\omega t) \\ \\ mu_p''(t) &= -m\omega^2A\cos(\omega t) - m\omega^2B\sin(\omega t) \\ \\ F_0\cos(\omega t) &= \left((k - m\omega^2)A + \gamma\omega B\right)\cos(2t) + \left((k - m\omega^2)B - \gamma\omega A\right)\sin(2t) \end{align*}$$
This allows us to define a system of two equations and two unknowns, write down the corresponding matrix equation, and solve it for $A$ and $B$

$$\begin{align*} (k - m\omega^2)A + \gamma\omega B &= F_0 \\ \\ -\gamma\omega A + (k - m\omega^2)B &= 0 \\ \\ \left[\begin{array}{cc} (k - m\omega^2) & \gamma\omega \\ -\gamma\omega & (k - m\omega^2) \end{array}\right]\left[\begin{array}{c} A \\ B \end{array}\right] &= \left[\begin{array}{c} F_0 \\ 0 \end{array}\right] \\ \\ \left|\begin{array}{cc} (k - m\omega^2) & \gamma\omega \\ -\gamma\omega & (k - m\omega^2) \end{array}\right| &= (k - m\omega^2)^2 + \gamma^2\omega^2 \\ \\ A = \dfrac{\left|\begin{array}{cc} F_0 & \gamma\omega \\ 0 & (k - m\omega^2) \end{array}\right|}{(k - m\omega^2)^2 + \gamma^2\omega^2} &= \dfrac{F_0(k - m\omega^2)}{(k - m\omega^2)^2 + \gamma^2\omega^2} \\ \\ B = \dfrac{\left|\begin{array}{cc} (k - m\omega^2) & F_0 \\ -\gamma\omega & 0 \end{array}\right|}{(k - m\omega^2)^2 + \gamma^2\omega^2} &= \dfrac{F_0\gamma\omega}{(k - m\omega^2)^2 + \gamma^2\omega^2} \end{align*}$$
Having found $A$ and $B$, we can determine $R^2 = A^2 + B^2$

$$\begin{align*} R^2 &= \left(\dfrac{F_0(k - m\omega^2)}{(k - m\omega^2)^2 + \gamma^2\omega^2}\right)^2 + \left(\dfrac{F_0\gamma\omega}{(k - m\omega^2)^2 + \gamma^2\omega^2}\right)^2 \\ \\ &= \dfrac{\sqrt{\left(F_0(k - m\omega^2)\right)^2 + \left(F_0\gamma\omega\right)^2}}{(k - m\omega^2)^2 + \gamma^2\omega^2} \\ \\ &= \dfrac{F_0\sqrt{(k - m\omega^2)^2 + (\gamma\omega)^2}}{(k - m\omega^2)^2 + \gamma^2\omega^2} \end{align*}$$
This allows us to find the steady state amplitude $R(m)$

Steady State Amplitude

$$ R(m) = \dfrac{F_0}{\sqrt{(k - m\omega^2)^2 + \gamma^2\omega^2}}, $$
Note: This equation will be referred to often in the rest of this section.

and compute its derivative with respect to mass $m$

$$\begin{align*} \dfrac{dR}{dm} &= -\dfrac{1}{2}F_0\left((k - m\omega^2)^2 + \gamma^2\omega^2\right)^{-3/2}\left(2(k - m\omega^2)\right)\left(-\omega^2\right) \\ \\ &= \dfrac{F_0\omega^2(k - m\omega^2)}{\left((k - m\omega^2)^2 + \gamma^2\omega^2\right)^{3/2}}. \end{align*}$$
Setting $\frac{dR}{dm} = 0$, we solve for the critical value of $m$

$$\begin{align*} F_0\omega^2(k - m\omega^2) &= 0 \\ \\ k - m\omega^2 &= 0 \\ \\ m\omega^2 &= k \\ \\ m &= \dfrac{k}{\omega^2} = \dfrac{16}{2^2} = 4\,\text{slugs}. \end{align*}$$
We now plug this value into $R(m)$ to determine the maximal amplitude of the steady state response

$$ R\left(\dfrac{k}{\omega^2}\right) = \dfrac{F_0}{\sqrt{(k - m\omega^2)^2 + \gamma^2\omega^2}} = \dfrac{F_0}{\gamma\omega} = \dfrac{4}{\frac{1}{4}\cdot 2} = 8.$$
We will verify that this is indeed a maximum by using the second derivative test . For simplicity, we plug in our values for $\gamma$ and $k$ into $\frac{dR}{dm}$ before we differentiate a second time

$$\begin{align*} \dfrac{dR}{dm} &= \dfrac{4\left(2^2\right)\left(16 - m\left(2^2\right)\right)}{\left((16 - 4m)^2 + \left(\frac{1}{4}\right)^2\left(2^2\right)\right)^{3/2}} \\ \\ &= \dfrac{16(16 - 4m)}{\left(16(4 - m)^2 + \frac{4}{16}\right)^{3/2}} = \dfrac{4-m}{\left((4-m)^2 + \frac{1}{64}\right)^{3/2}} \\ \\ &= (4-m)\left((m-4)^2 + \frac{1}{64}\right)^{-3/2} \end{align*}$$
$$\begin{align*} \dfrac{d^2R}{dm^2} &= -m\left((m-4)^2 + \frac{1}{64}\right)^{-3/2} \\ &\ \qquad - (m-4)\left(-\dfrac{3}{2}\right)\left((m-4)^2 + \frac{1}{64}\right)^{-5/2}(2)(m-4) \\ \\ &= -\dfrac{m}{\left((m-4)^2 + \frac{1}{64}\right)^{3/2}} + \dfrac{3(m-4)^2}{\left((m-4)^2 + \frac{1}{64}\right)^{5/2}} \end{align*}$$
Lastly, we plug in our mass $m=4$ to check that $\frac{d^2R}{dm^2} \lt 0$,

$$ \dfrac{d^2R}{dm^2}(4) = -\dfrac{4}{\left(0 + \frac{1}{64}\right)^{3/2}} + 0 = -4\left(8\right)^3 = -2048 \lt 0.$$
Thus $m=4$, $R=8$ is a maximum.

Amplitude as a Function of Mass

3.8.4 Steady State Amplitude as a Function of Frequency

It is also possible to maximize steady state amplitude $R$ with respect to the frequency $\omega$. Recalling from our previous example that

$$\begin{align*} R &= \dfrac{F_0}{\sqrt{(k - m\omega^2)^2 + \gamma^2\omega^2}} \\ \\ &= \dfrac{F_0}{k\sqrt{\left(1-\frac{m}{k}\omega^2\right)^2 + \left(\frac{\gamma\omega}{k}\right)^2}}, \\ \end{align*}$$
we may write

$$R = \dfrac{F_0}{k}\left(\left(1-\frac{m}{k}\omega^2\right)^2 + \left(\frac{\gamma\omega}{k}\right)^2\right)^{-1/2},$$
which we will use both here and later in this subsection. Differentiating this expression with respect to frequency of the forcing function $\omega$ gives

$$\begin{align*} \dfrac{dR}{d\omega} &= \dfrac{F_0}{k}\left(-\dfrac{1}{2}\right)\left(\left(1-\frac{m}{k}\omega^2\right)^2 + \left(\frac{\gamma\omega}{k}\right)^2\right)^{-3/2}\left(2\left(1-\frac{m}{k}\omega^2\right)\left(\dfrac{-2m\omega}{k}\right)+\dfrac{2\gamma^2\omega}{k^2}\right) \\ \\ &= -\dfrac{F_0\omega}{k^3}\left(\gamma^2 - 2mk\left(1-\frac{m}{k}\omega^2\right)\right)\left(\left(1-\frac{m}{k}\omega^2\right)^2 + \left(\frac{\gamma\omega}{k}\right)^2\right)^{-3/2} \\ \end{align*}$$
Setting $\frac{dR}{d\omega}=0$, we solve for the frequency $\omega_{\text{max}}$ that maximizes $R$

$$\begin{align*} \gamma^2 - 2mk\left(1-\dfrac{m}{k}\omega^2\right) &= 0 \\ \\ 2mk\left(1-\dfrac{m}{k}\omega^2\right) &= \gamma^2 \\ \\ 1-\dfrac{m}{k}\omega^2 &= \dfrac{\gamma^2}{2mk} \\ \\ \dfrac{m}{k}\omega^2 &= 1 - \dfrac{\gamma^2}{2mk} \\ \\ \omega_{\text{max}}^2 &= \dfrac{k}{m}\left(1 - \dfrac{\gamma^2}{2mk}\right) \\ \\ \end{align*}$$
Note that the natural frequency is

$$\omega_0 = \sqrt{\dfrac{k}{m}}.$$
Substituting $\omega_0$ into our equation for $\omega_{\text{max}}$, we obtain

$$\omega_{\text{max}}^2 = \omega_0^2\left(1 - \dfrac{\gamma^2}{2mk}\right) = \omega_0^2 - \dfrac{\gamma^2}{2m^2}.$$
Now substituting this into our earlier expression for $R$, the maximum value for the amplitude is

$$\begin{align*} R_{\text{max}} &= \dfrac{F_0}{\sqrt{\left(k - m\left(\omega_0^2 - \dfrac{\gamma^2}{2m^2}\right)\right)^2 + \gamma^2\left(\omega_0^2 - \dfrac{\gamma^2}{2m^2}\right)}} \\ \\ &= \dfrac{F_0}{\sqrt{\left(k - k + \dfrac{\gamma^2}{2m}\right)^2 + \gamma^2\omega_0^2 - \gamma^2\dfrac{\gamma^2}{2m^2}}} \\ \\ &= \dfrac{F_0}{\sqrt{\gamma^2\dfrac{\gamma^2}{4m^2} + \gamma^2\omega_0^2 - \gamma^2\dfrac{\gamma^2}{2m^2}}} \\ \\ &= \dfrac{F_0}{\gamma\sqrt{\dfrac{\gamma^2}{4m^2} + \omega_0^2 - \dfrac{\gamma^2}{2m^2}}} = \dfrac{F_0}{\gamma\sqrt{\omega_0^2 - \dfrac{\gamma^2}{4m^2}}} \\ \\ &= \dfrac{F_0}{\gamma\sqrt{\omega_0^2 - \dfrac{k\gamma^2}{4m^2k}}} = \dfrac{F_0}{\gamma\sqrt{\omega_0^2 - \omega_0^2\dfrac{\gamma^2}{4mk}}} \\ \\ &= \dfrac{F_0}{\gamma\omega_0\sqrt{1 - \dfrac{\gamma^2}{4mk}}} \\ \\ \end{align*}$$
Setting $\omega_0^2 = \frac{k}{m}$ that same starting expression for $R$ results in

$$\begin{align*} R(\omega) &= \dfrac{F_0}{k}\left(\left(1-\frac{m}{k}\omega^2\right)^2 + \left(\frac{\gamma\omega}{k}\right)^2\right)^{-1/2} \\ \\ &= \dfrac{F_0}{k}\left(\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2 + \left(\frac{m\gamma^2\omega^2}{mk^2}\right)\right)^{-1/2} \\ \\ &= \dfrac{F_0}{k}\left(\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2 + \left(\dfrac{\gamma^2}{mk}\right)\left(\frac{m\omega^2}{k}\right)\right)^{-1/2} \\ \\ &= \dfrac{F_0}{k}\left(\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2 + \Gamma\left(\frac{\omega}{\omega_0}\right)^2\right)^{-1/2} \end{align*}$$
where

$$\Gamma = \frac{\gamma^2}{mk}.$$
For a fixed $m$ and $k$, and with relatively small $\gamma$, $\Gamma$ is small as well, hence

$$R_{\text{max}} = \dfrac{F_0}{\gamma\omega_0\sqrt{1 - \dfrac{\gamma^2}{4mk}}} = \dfrac{F_0}{\gamma\omega_0\sqrt{1 - \dfrac{\Gamma}{4}}} \approx \dfrac{F_0}{\gamma\omega_0}.$$
Letting $\gamma\rightarrow 0$ for fixed $m$ and $k$ means that $\Gamma\rightarrow 0$ also and so the amplitude

$$\lim_{\gamma\rightarrow 0}R = \lim_{\gamma\rightarrow 0} \dfrac{F_0}{\gamma\omega_0\sqrt{1 - \dfrac{\gamma^2}{4mk}}} = +\infty.$$
This means that if the frequency of the forcing function is close to the natural frequency of the steady state response, the amplitude can grow without bound for lightly damped systems (small $\gamma$). The phenomenon is called resonance and will be discussed more shortly.

3.8.5 Pay Attention to the Units

Recall our primary mathematical model

$$mu'' + \gamma u' + ku = F_0\cos(\omega t),$$
with parameters $\omega_0^2 = \dfrac{k}{m}$, $\Gamma = \dfrac{\gamma^2}{mk}$, and

$$ R(\omega) = \dfrac{F_0}{k}\left(\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2 + \Gamma\left(\frac{\omega}{\omega_0}\right)^2\right)^{-1/2}, $$
which may be rearranged as

$$ \dfrac{Rk}{F_0} = \left(\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2 + \Gamma\left(\frac{\omega}{\omega_0}\right)^2\right)^{-1/2}.$$
Also, recall our general solution

$$u(t) = y(t) = e^{-\sigma t}\left(c_1\cos\left(\mu t\right) + c_2\sin\left(\mu t\right)\right) + R\cos\left(\omega t - \delta\right),$$

where

$$\begin{align*} \sigma &= \dfrac{\gamma}{2m} \\ \\ \mu &= \sqrt{\frac{k}{m}-\sigma^2} = \sqrt{\omega_0^2 - \sigma^2} \\ \\ A &= \dfrac{F_0(k - m\omega^2)}{(k - m\omega^2)^2 + \gamma^2\omega^2} \\ \\ B &= \dfrac{F_0\gamma\omega}{(k - m\omega^2)^2 + \gamma^2\omega^2} \\ \\ \tan(\delta) &= \dfrac{k - m\omega^2}{\gamma\omega}. \end{align*}$$
If we consider our units, we have

$$\begin{align*} R,\ c_1,\ c_2 &= \text{displacement} = \text{ft} \\ \\ F_0 &= \text{force} = \dfrac{\text{slug}\cdot\text{ft}}{\text{s$^2$}} = \text{lb} \\ \\ k &= \text{spring constant} = \dfrac{\text{force}}{\text{displacement}} = \dfrac{\text{lb}}{\text{ft}} \\ \\ \gamma &= \frac{\text{force}}{\text{velocity}} = \dfrac{\text{lb}}{\text{ft/s}} = \dfrac{\text{slug}\cdot\text{ft}}{\text{s$^2$}}\cdot\dfrac{\text{s}}{\text{ft}} = \dfrac{\text{slug}}{\text{s}} \\ \\ \omega_0 &= \sqrt{\dfrac{\text{lb/ft}}{\text{mass}}} = \sqrt{\dfrac{\dfrac{\text{slug}\cdot\text{ft}}{\text{s$^2$}}\cdot\dfrac{1}{\text{ft}}}{\text{slug}}} = \sqrt{\dfrac{1}{\text{s$^2$}}} = \dfrac{1}{\text{s}} = \text{hertz} \\ \\ \omega &= \text{hertz} \\ \\ \sigma &= \dfrac{\dfrac{\text{slug}}{\text{s}}}{\text{slug}} = \dfrac{1}{\text{s}} = \text{hertz} \\ \\ \Gamma &= \frac{\gamma^2}{mk} = \dfrac{\gamma}{m}\cdot\dfrac{\gamma}{k} = \dfrac{1}{s}\cdot\dfrac{\dfrac{\text{lb}}{\text{ft/s}}}{\dfrac{\text{lb}}{\text{ft}}} = \dfrac{1}{s}\cdot\dfrac{\dfrac{1}{s}}{1} = 1 \\ \\ \dfrac{Rk}{F_0} &= \dfrac{\text{ft}\cdot\dfrac{\text{lb}}{\text{ft}}}{\text{lb}} = 1 \end{align*}$$

3.8.6 Resonance

The work in thee previous subsection shows that $\dfrac{Rk}{F_0}$ and $\Gamma$ are unitless. If we plot the response amplitude of the steady state solution $\dfrac{Rk}{F_0}$ versus the ratio $\dfrac{\omega}{\omega_0}$ for various values of $\Gamma$


we see that as $\omega$ nears $\omega_0$; that is $\dfrac{\omega}{\omega_0}$ approaches $1$, the response amplitude grows significantly when $\Gamma = \dfrac{\gamma^2}{mk}$ is small.

This does not imply that the resistance (drag or friction) must be small. For example a very stiff system, one with a high value for $k$, or a large mass relative to $\gamma$ will still result is a small damping parameter $\Gamma$. In these systems the amplitude of the forced response will be quite large when $\omega$ is near $\omega_0$. This is the cause of resonance .

3.8.7 Steady State Amplitude as a Function of Capacitance

In a RLC circuit problem we instead use the model

$$Lq'' + Rq' + \dfrac{1}{C}q = F_0\cos(\omega t).$$
Since $R$ is the ( Thevenin equivalent ) resistance of the circuit we will rename the amplitude of the steady state response $A$ and its equation becomes

$$A = \dfrac{CF_0}{\sqrt{\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2}}.$$
To obtain the maximum response amplitude with respect to the capacitance $C$, we differentiate

$$\begin{align*} \dfrac{dA}{dC} &= F_0\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{-1/2} \\ &\qquad -\left(\dfrac{1}{2}\right)CF_0\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{-3/2}\left(2\left(1 - L\omega^2C\right)\left(-L\omega^2\right) + 2R^2\omega^2C\right) \\ \\ &= \dfrac{F_0}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{1/2}} - \dfrac{CF_0\left(-L\omega^2 + L^2\omega^4C + R^2\omega^2C\right)}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{3/2}} \\ \\ &= \dfrac{F_0\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{3/2}} - \dfrac{F_0\left(-L\omega^2C + L^2\omega^4C^2 + R^2\omega^2C^2\right)}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{3/2}} \\ \\ &= \dfrac{F_0\left(\left(1 - 2L\omega^2C + L^2\omega^4C^2 + R^2\omega^2C^2\right)-\left(-L\omega^2C + L^2\omega^4C^2 + R^2\omega^2C^2\right)\right)}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{3/2}} \\ \\ &= \dfrac{F_0\left(1 - L\omega^2C\right)}{\left(\left(1 - LC\omega^2\right)^2 + \left(RC\omega\right)^2\right)^{3/2}}. \end{align*}$$
We only need the numerator to equal zero and $F_0$ is a non-zero constant, so

$$\begin{align*} 1 - L\omega^2C &= 0 \\ \\ L\omega^2C &= 1 \\ \\ C_{\text{max}} &= \dfrac{1}{L\omega^2}. \end{align*}$$
Substituting this expression for $C$ into our equation from the steady state amplitude $A$,

$$\begin{align*} A_{\text{max}} &= \dfrac{\dfrac{F_0}{L\omega^2}}{\left(\left(1 - \dfrac{L\omega^2}{L\omega^2}\right)^2 + \left(\dfrac{R\omega}{L\omega^2}\right)^2\right)^{1/2}} \\ \\ &= \dfrac{\dfrac{F_0}{L\omega^2}}{\dfrac{R}{L\omega}} = \dfrac{F_0}{L\omega^2}\cdot\dfrac{L\omega}{R} = \dfrac{F_0}{R\omega}. \end{align*}$$
So our maximal amplitude as a function of capacitance $A_\text{max} = \dfrac{F_0}{R\omega}$.

3.8.8 Envelopes, Beats, and Resonance

Example 3.8.3

Consider the vibrating system described by the differential equation

$$u'' + u = 3\cos\left(\omega t\right),\qquad u(0)=0,\ u'(0)=0.$$
Find the solution for $\omega\neq 1$, and graph the solution $u(t)$ vs $t$ for $\omega=0.7$, $\omega=0.8$, and $\omega=0.9$ on the same plot. Describe how the response varies as $\omega$ varies in this interval. What happens as $\omega$ gets closer and closer to the natural frequency $1$?

Solution

The characteristic equation is $d^2 + 1 = 0$ and has roots $d = \pm i$. Hence the solution of the associated homogeneous problem is

$$u_h(t) = c_1\cos(t) + c_2\sin(t).$$
We find the particular solution using undetermined coefficients when $\omega\neq 1$. Beginning by stating $u_p(t)$ and computing its derivatives,

$$\begin{align*} u_p(t) &= A\cos(\omega t) + B\sin(\omega t) \\ \\ u_p'(t) &= B\omega\cos(\omega t) - A\omega\sin(\omega t) \\ \\ u_p''(t) &= -A\omega^2\cos(\omega t) - B\omega^2\sin(\omega t). \end{align*}$$
We may now construct a system of two equations and two unknowns by plugging these values into the ODE and collecting terms

$$ A\left(1-\omega^2\right)\cos(\omega t) + B\left(1-\omega^2\right)\sin(\omega t) = 3\cos(\omega t)$$
which yields the equations

$$\begin{matrix} A\left(1-\omega^2\right) = 3 & \qquad & B\left(1-\omega^2\right) = 0 \end{matrix}$$
whose solutions are

$$\begin{matrix} A = \dfrac{3}{1-\omega^2} & \quad\qquad & B = 0 \end{matrix}$$
This means that the general solution of the nonhomogeneous equation is given by

$$ u(t) = c_1\cos(t) + c_2\sin(t) + \dfrac{3}{1-\omega^2}\cos(\omega t) $$
and its derivative is

$$ u'(t) = -c_1\sin(t) + c_2\cos(t) - \dfrac{3\omega}{1-\omega^2}\sin(\omega t).$$
We may find the values $c_1$ and $c_2$ by applying the initial conditions to these expressions.

You must include the forcing function in your computation of the coefficients.

$$\begin{matrix} u(0) = c_1 + 0 + \dfrac{3}{1-\omega^2} = 0 & \qquad & c_1 = -\dfrac{3}{1-\omega^2} \\ \\ u'(0) = 0 + c_2 - 0 = 0 & \qquad & c_2 = 0 \end{matrix}$$

So the solution to the initial value problem is given by

$$ u(t) = \dfrac{3}{1-\omega^2}\left(\cos(\omega t) - \cos(t)\right). $$
Since this expression features a difference of cosine functions, it is possible to rewrite it using trigonometric identities.

$$\begin{align*} \cos(u - v) &= \cos(u)\cos(v) + \sin(u)\sin(v) \\ \\ \cos(u + v) &= \cos(u)\cos(v) - \sin(u)\sin(v) \\ \\ \cos(u-v) - \cos(u+v) &= 2\sin(u)\sin(v) \end{align*}$$
We consider $u + v = t$ and $u - v = \omega t$, which we may add to see

$$\begin{align*} 2u &= \omega t + t \\ \\ u &= \dfrac{1 + \omega}{2}t, \end{align*}$$
and subtract to obtain

$$\begin{align*} 2v &= \omega t - t \\ \\ v &= \dfrac{1 - \omega}{2}t. \end{align*}$$
At last, we rewrite our formula for the solution to the initial value problem using this new expression

$$ u(t) = \dfrac{6}{1-\omega^2}\sin\left(\dfrac{1 + \omega}{2}t\right)\sin\left(\dfrac{1 - \omega}{2}t\right). $$


The solution exhibits amplitude modulation. The envelope increases in amplitude and period as $\omega$ gets closer to $1$ from below, while the beat frequency increases. As $\omega$ gets closer and closer to the natural frequency $1$, the envelope period and amplitude will grow without bound resulting in resonance .

Exercise 3.8.1

Consider the vibrating system described by the differential equation

$$u'' + u = 3\cos\left(\omega t\right),\qquad u(0) = 1,\ u'(0) = 1.$$
Find the solution for $\omega\neq 1$, and plot the solution $u(t)$ vs $t$ for $\omega=0.7$, $\omega=0.8$, and $\omega=0.9$. Compare the results with the solution in Example 3 .

View Solution

The general solution is still
$$u(t) = c_1\cos(t) + c_2\sin(t) + \dfrac{3}{1-\omega^2}\cos(\omega t),$$
so we need only to recompute $c_1$ and $c_2$ with our new initial conditions
$$\begin{matrix} u(0) = c_1 + 0 + \dfrac{3}{1-\omega^2} = 1 & \qquad & c_1 = 1 -\dfrac{3}{1-\omega^2} \\ \\ u'(0) = 0 + c_2 - 0 = 1 & \qquad & c_2 = 1. \end{matrix}$$
Therefore, the solution to the IVP is
$$\begin{align*} u(t) &= \left(1 - \dfrac{3}{1-\omega^2}\right)\cos(t) + \sin(t) + \dfrac{3}{1-\omega^2} \cos(\omega t) \\ \\ &= \cos(t) + \sin(t) + \dfrac{3}{1-\omega^2}\left(\cos(\omega t) - \cos(t)\right). \end{align*}$$
For convenience, we would like to convert homogeneous part of the solution $\cos\omega t + \sin\omega t$ to the form $R\cos\left(\omega t - \delta\right)$. Solving for $R$ and $\delta$,
$$\begin{matrix} R = \sqrt{1^2 + 1^2} & \qquad & \tan\delta = \dfrac{1}{1} \\ \\ R = \sqrt{2} & \qquad & \delta = \dfrac{\pi}{4}. \end{matrix}$$
So our solution is
$$ u(t) = \sqrt{2}\cos\left(t-\frac{\pi}{4}\right) + \dfrac{6}{1-\omega^2}\sin\left(\dfrac{1 + \omega}{2}t\right)\sin\left(\dfrac{1 - \omega}{2}t\right) $$
We are comparing this solution to the solution of Example 3.8.3

$$u_{\text{old}}(t) = \dfrac{6}{1-\omega^2}\sin\left(\dfrac{1 + \omega}{2}t\right)\sin\left(\dfrac{1 - \omega}{2}t\right),$$
and see that
$$u(t) = \sqrt{2}\cos\left(t-\frac{\pi}{4}\right) + u_{\text{old}}(t).$$
This will result is a slightly different curve for each forcing frequency $\omega$. For example, with $\omega=0.7$

If we graph the three solutions for $\omega=0.7$, $\omega=0.8$, and $\omega=0.9$, the solutions will look very similar and the result will be the same as $\omega$ get closer and closer to the natural frequency $1$.

Interactive Example

Here is an interactive Desmos example that allows you explore the effect of $\omega\rightarrow 1$ in our problem.

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