Unit D5: Mechanical and Electrical Vibrations机械振动与电学振动

A $2\ \text{kg}$ mass stretches a spring $0.5\ \text{m}$ at rest under gravity. The damping force is $8$ newtons when the speed is $4\ \text{m/s}$. Write the equation of motion with no external force. Take $g = 10\ \text{m/s}^2$.

Hooke's law at equilibrium: $mg = k\,x_0$, so $k = \dfrac{(2)(10)}{0.5} = 40\ \text{N/m}$.

Damping is linear: $c\,(4) = 8$, so $c = 2\ \text{N}\cdot\text{s/m}$.

Thus $\omega_0 = \sqrt{20} \approx 4.47\ \text{rad/s}$ and $\beta = \tfrac{1}{2}$.

A $0.25\ \text{kg}$ mass hangs from a spring and stretches it $0.05\ \text{m}$ at rest. There is no damping. Measuring $y$ downward from the natural (unstretched) length, Newton's law reads $m y'' = mg - k y$. Show that measuring $x$ from the equilibrium position removes the gravity term, and write the resulting equation. Take $g = 9.8\ \text{m/s}^2$.

At equilibrium the spring force balances gravity: $k\,y_0 = mg$, so $k = \dfrac{(0.25)(9.8)}{0.05} = 49\ \text{N/m}$. Now substitute $y = y_0 + x$, where $x$ is the displacement from equilibrium. Since $y_0$ is constant, $y'' = x''$, and

The constant gravitational offset is absorbed exactly by the new equilibrium, leaving $m x'' + k x = 0$. Numerically $0.25\,x'' + 49\,x = 0$, so $\omega_0 = \sqrt{49/0.25} = \sqrt{196} = 14\ \text{rad/s}$.

Given $3x'' + 12x' + 75x = 6\cos 4t$, identify $m$, $c$, $k$, $\omega_0$, $\beta$, the damping ratio $\zeta$, and the driving frequency.

Read coefficients directly: $m = 3$, $c = 12$, $k = 75$. Then

The damping ratio is $\zeta = \beta/\omega_0 = 2/5 = 0.4 < 1$, so the free system is underdamped. The drive $6\cos 4t$ has driving frequency $\omega = 4$, which is below $\omega_0 = 5$. Dividing through by $m=3$ puts it in standard form: $x'' + 4x' + 25x = 2\cos 4t$.

Solve $x'' + 16x = 0$ with $x(0) = 3$, $x'(0) = -8$, and report the amplitude and period.

Here $\omega_0 = 4$. The general solution is $x = c_1\cos 4t + c_2\sin 4t$.

From $x(0)=3$: $c_1 = 3$. From $x'(t) = -4c_1\sin 4t + 4c_2\cos 4t$ and $x'(0)=-8$: $4c_2 = -8$, so $c_2 = -2$.

Write $x(t) = 3\cos 4t - 2\sin 4t$ from the previous example in the single-cosine form $A\cos(\omega_0 t - \phi)$.

Match $A\cos(\omega_0 t - \phi) = A\cos\phi\,\cos\omega_0 t + A\sin\phi\,\sin\omega_0 t$ against $3\cos 4t - 2\sin 4t$. So $A\cos\phi = 3$ and $A\sin\phi = -2$, giving

Because $\cos\phi > 0$ and $\sin\phi < 0$, the phase $\phi$ lies in the fourth quadrant: $\phi = \arctan(-2/3) \approx -0.588$ radians. Hence $x(t) = \sqrt{13}\,\cos(4t + 0.588)$. The amplitude $\sqrt{13}$ matches the value computed from $\sqrt{c_1^2 + c_2^2}$, a useful self-check.

Define the total energy $E = \tfrac12 m (x')^2 + \tfrac12 k x^2$, the sum of kinetic and elastic potential energy. Differentiate along a solution of $mx'' + kx = 0$:

So $E$ is constant in time. The motion never decays, which is exactly why the undamped solution is a pure sinusoid of fixed amplitude. Reintroducing $c>0$ gives $dE/dt = -c(x')^2 \le 0$, the energy loss that produces damping.

Solve $x'' + 4x' + 3x = 0$ with $x(0)=1$, $x'(0)=1$, and classify the damping.

Characteristic equation: $r^2 + 4r + 3 = (r+1)(r+3)=0$, roots $r=-1, -3$. Two distinct negative reals, so the system is overdamped.

Apply data: $c_1 + c_2 = 1$ and $-c_1 - 3c_2 = 1$. Subtracting gives $-2c_2 = 2$, so $c_2 = -1$ and $c_1 = 2$.

Solve $x'' + 2x' + 5x = 0$ with $x(0)=4$, $x'(0)=0$.

Roots: $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$, so $\beta = 1$ and $\omega_1 = 2$.

From $x(0)=4$: $c_1 = 4$. Then $x'(0) = -c_1 + 2c_2 = 0$ gives $c_2 = 2$.

The envelope has amplitude $e^{-t}\sqrt{4^2 + 2^2} = e^{-t}\sqrt{20} = 2\sqrt5\,e^{-t}$, and the quasi-frequency is $\omega_1 = 2$, slightly below the natural frequency $\omega_0 = \sqrt5 \approx 2.236$.

Solve $x'' + 6x' + 9x = 0$ with $x(0) = 2$, $x'(0) = -3$, and describe the motion.

Characteristic equation: $r^2 + 6r + 9 = (r+3)^2 = 0$, a repeated root $r = -3$. The discriminant $c^2 - 4mk = 36 - 36 = 0$ confirms critical damping. The general solution carries the telltale factor of $t$:

From $x(0) = 2$: $c_1 = 2$. Differentiate using the product rule, $x'(t) = c_2 e^{-3t} - 3(c_1 + c_2 t)e^{-3t}$, so $x'(0) = c_2 - 3c_1 = c_2 - 6 = -3$, giving $c_2 = 3$.

Because $2 + 3t > 0$ for all $t \ge 0$, the mass never crosses equilibrium: it eases back to rest without a single oscillation. This is the fastest possible non-oscillatory return, which is exactly why shock absorbers and door closers are tuned near critical damping.

Fix $\omega_0$ and ask which damping rate $\beta$ makes the free response decay to zero quickest. The slowest-decaying piece of the solution sets the long-run rate, because as $t \to \infty$ that term dominates.

Overdamped ($\beta > \omega_0$). The roots are $r_\pm = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$, both negative. The slower one is $r_+ = -\beta + \sqrt{\beta^2 - \omega_0^2}$. Write $s = \sqrt{\beta^2 - \omega_0^2} > 0$; then $r_+ = -\beta + s$. Since $s = \sqrt{\beta^2 - \omega_0^2} > \sqrt{\beta^2 - \omega_0^2 - \text{(anything)}}$, as $\beta$ increases past $\omega_0$ the quantity $\beta - s = \beta - \sqrt{\beta^2 - \omega_0^2} = \dfrac{\omega_0^2}{\beta + \sqrt{\beta^2-\omega_0^2}}$ decreases toward $0$. So heavier damping makes $|r_+|$ smaller, meaning the dominant term decays slower, not faster.

Underdamped ($\beta < \omega_0$). The envelope is $e^{-\beta t}$, whose rate is exactly $\beta < \omega_0$.

Critical ($\beta = \omega_0$). The decay rate is $\beta = \omega_0$, the largest value reachable before the overdamped branch starts slowing down. The factor $t$ in $(c_1 + c_2 t)e^{-\omega_0 t}$ is subdominant to the exponential, so the effective rate is still $\omega_0$. Therefore the decay rate is maximized exactly at $\beta = \omega_0$, the critical case.

Find the steady-state solution of $x'' + 2x' + 10x = 4\cos 3t$.

Try $x_p = a\cos 3t + b\sin 3t$. Then $x_p'' = -9x_p$, so substituting gives

Match coefficients: $\cos$: $a + 6b = 4$; $\sin$: $b - 6a = 0$, so $b = 6a$. Then $a + 36a = 4$ gives $a = \tfrac{4}{37}$, $b = \tfrac{24}{37}$.

For $x'' + \omega_0^2 x = \tfrac{F_0}{m}\cos\omega_0 t$, the forcing frequency $\omega_0$ already solves the homogeneous equation, so $\cos\omega_0 t$ is a root of the operator. The method of undetermined coefficients then requires multiplying the trial solution by $t$. Substituting $x_p = t(a\cos\omega_0 t + b\sin\omega_0 t)$ and computing $x_p''$:

Matching to $\tfrac{F_0}{m}\cos\omega_0 t$ gives $a = 0$ and $b = \tfrac{F_0}{2m\omega_0}$, hence $x_p = \tfrac{F_0}{2m\omega_0}\,t\sin\omega_0 t$. The $t$ in front is what drives the amplitude to infinity.

Solve $x'' + 2x' + 10x = 4\cos 3t$ with $x(0) = 0$, $x'(0) = 0$, and identify the transient and steady-state parts.

The steady state was found in Example 4.1: $x_p = \tfrac{4}{37}\cos 3t + \tfrac{24}{37}\sin 3t$. For the homogeneous part, $r^2 + 2r + 10 = 0$ gives $r = -1 \pm 3i$, so

Apply $x(0) = 0$: $c_1 + \tfrac{4}{37} = 0$, so $c_1 = -\tfrac{4}{37}$. Differentiate the full solution and set $t=0$. Using $x'(0) = 0$ and $x_p'(0) = \tfrac{72}{37}$, while $x_h'(0) = -c_1 + 3c_2$,

So $x(t) = e^{-t}\!\left(-\tfrac{4}{37}\cos 3t - \tfrac{76}{111}\sin 3t\right) + \tfrac{4}{37}\cos 3t + \tfrac{24}{37}\sin 3t$. The first group decays (the transient); the second group is the persistent steady state.

An $LC$-series circuit has $L = 1\ \text{H}$, $R = 4\ \Omega$, $C = \tfrac14\ \text{F}$, and no source. Find $Q(t)$ given $Q(0)=2$, $Q'(0)=0$.

The equation is $Q'' + 4Q' + 4Q = 0$ since $1/C = 4$. Characteristic: $r^2 + 4r + 4 = (r+2)^2 = 0$, a repeated root $r=-2$. The circuit is critically damped.

From $Q(0)=2$: $c_1 = 2$. Then $Q'(t) = (c_2 - 2c_1 - 2c_2 t)e^{-2t}$, so $Q'(0) = c_2 - 4 = 0$, giving $c_2 = 4$.

A series circuit has $L = 0.5\ \text{H}$, $R = 6\ \Omega$, and $C = 0.02\ \text{F}$, with no source. Classify the circuit and find its quasi-frequency.

Here $1/C = 50$, so the equation in the charge is $0.5\,Q'' + 6\,Q' + 50\,Q = 0$, or in standard form $Q'' + 12\,Q' + 100\,Q = 0$. Then $\omega_0 = \sqrt{1/(LC)} = \sqrt{1/(0.5 \cdot 0.02)} = \sqrt{100} = 10$ and $\beta = R/(2L) = 6/1 = 6$.

Classify by comparing $R^2 = 36$ with $4L/C = 4(0.5)(50) = 100$. Since $R^2 < 4L/C$, the circuit is underdamped. The quasi-frequency is

The charge therefore oscillates as $Q(t) = A\,e^{-6t}\cos(8t - \phi)$, an electrical analogue of the underdamped spring of Example 3.2.

Kirchhoff's voltage law states that around a closed loop the source voltage equals the sum of the voltage drops. For a series circuit the three passive elements contribute, in terms of the current $I$ and the charge $Q$:

The inductor law $V_L = L\,dI/dt$ comes from Faraday's law; the resistor law $V_R = RI$ is Ohm's law; and the capacitor stores charge so its voltage is $Q/C$. Summing and equating to the source $E(t)$,

Now use $I = dQ/dt$, so $dI/dt = d^2Q/dt^2 = Q''$. Substituting turns the loop equation into a second-order ODE in $Q$ alone:

Term by term this is identical in form to $m x'' + c x' + k x = F(t)$. The correspondence is not a coincidence of notation: inductance opposes change in current the way mass opposes change in velocity, resistance dissipates energy the way friction does, and the capacitor stores energy the way a stretched spring does.

For $x'' + 0.4x' + 25x = \cos\omega t$, find the driving frequency that maximizes the steady-state amplitude.

Here $\omega_0^2 = 25$ and $2\beta = 0.4$, so $\beta = 0.2$ and $\beta^2 = 0.04$. Since $\beta^2 < \tfrac12(25)$, an interior peak exists.

The peak sits just below $\omega_0 = 5$, as expected for light damping.

The steady-state amplitude is $C(\omega) = \dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}}$. Maximizing $C$ is the same as minimizing the radicand $g(\omega) = (\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2$. Differentiate with respect to $\omega$:

Setting $g'(\omega)=0$ with $\omega>0$ gives $\omega^2 = \omega_0^2 - 2\beta^2$, hence $\omega_{\text{res}} = \sqrt{\omega_0^2 - 2\beta^2}$ whenever the radicand is positive.

Solve $x'' + 4x = \cos 1.8t$ with $x(0) = 0$, $x'(0) = 0$, and exhibit the beat envelope.

The natural frequency is $\omega_0 = 2$ and the drive frequency is $\omega = 1.8$, close but not equal, so expect beats. The particular solution is $x_p = \dfrac{1}{\omega_0^2 - \omega^2}\cos\omega t = \dfrac{1}{4 - 3.24}\cos 1.8t = \dfrac{1}{0.76}\cos 1.8t$. The homogeneous part is $c_1\cos 2t + c_2\sin 2t$.

From $x(0)=0$: $c_1 + \tfrac{1}{0.76} = 0$, so $c_1 = -\tfrac{1}{0.76}$. From $x'(0)=0$: $2c_2 = 0$, so $c_2 = 0$. Hence

That is $x(t) = \dfrac{2}{0.76}\,\sin(0.1t)\,\sin(1.9t)$: a fast carrier at $1.9\ \text{rad/s}$ inside a slow envelope $\sin(0.1t)$. The envelope period is $2\pi/0.1 \approx 62.8\ \text{s}$ between zeros, so the amplitude swells and fades slowly while the mass vibrates rapidly inside it.

For $x'' + 2x' + 5x = 3\cos t$, find the steady-state amplitude and phase lag using the complex gain.

Here $\omega_0^2 = 5$, $2\beta = 2$ so $\beta=1$, $\omega = 1$, and $F_0/m = 3$. The denominator is $(5-1) + 2i(1)(1) = 4 + 2i$.

The phase lag is $\delta = \arg(4 + 2i) = \arctan(2/4) = \arctan(0.5) \approx 0.4636$ radians.

For a lightly damped oscillator $x'' + 0.2x' + 9x = \cos\omega t$, compute the phase lag $\delta$ when the drive is tuned exactly to $\omega = \omega_0 = 3$, and confirm it is $90$ degrees.

Here $\omega_0^2 = 9$ and $2\beta = 0.2$. At $\omega = 3$ the complex denominator is

So $z = \dfrac{1}{0.6\,i} = -\dfrac{i}{0.6}$, a purely imaginary number pointing along the negative imaginary axis. Its argument is $-\pi/2$, so the phase lag is

At the natural frequency the response always lags the drive by exactly a quarter cycle, regardless of how small the damping is. The amplitude there is $C = |z| = 1/0.6 \approx 1.667$, the resonant peak.

Claim: if $x_p = \operatorname{Re}\!\big(z\,e^{i\omega t}\big)$ with $z = |z|\,e^{-i\delta}$, then $x_p = |z|\cos(\omega t - \delta)$, so $C = |z|$ and the lag is $\delta = -\arg z$. The derivation makes the complex shortcut honest.

Substitute $x_p = z\,e^{i\omega t}$ into the operator $L[x] = x'' + 2\beta x' + \omega_0^2 x$. Each derivative of $e^{i\omega t}$ brings down a factor $i\omega$:

Setting this equal to the complexified drive $\tfrac{F_0}{m}e^{i\omega t}$ and cancelling $e^{i\omega t}$ gives $z = \dfrac{F_0/m}{(\omega_0^2 - \omega^2) + 2i\beta\omega}$. Writing $z = |z|e^{-i\delta}$ in polar form,

Taking the modulus of $z$ reproduces $C = \dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}}$, exactly the steady-state amplitude derived by real methods in Section 4. The two approaches agree, but the complex one replaces a coefficient-matching system with a single division.