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Differential Equations · Section 3.7 · generated practice set

Mass, spring and damper: natural frequency from a hanging stretch, amplitude and phase, and the three damping cases.

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Method

A mass on a spring with damping obeys $$mu'' + gamma u' + ku = 0,$$ with $m$ the mass, $gamma ge 0$ the damping coefficient and $k$ the spring constant. It is the constant-coefficient second-order equation from 3.1–3.4 wearing physical clothes.

Getting $k$ from a hanging mass

If the mass stretches the spring by $L$ at rest, the forces balance: $$kL = mg quadRightarrowquad k = rac{mg}{L}.$$ Then $omega_0 = sqrt{k/m} = sqrt{g/L}$ — the mass cancels. Watch units: $L$ in metres with $g = 9.8$, or inches with $g = 32$ ft/s².

Undamped motion

$u'' + omega_0^2u = 0$ gives $u = c_1cosomega_0t + c_2sinomega_0t$, which is the single oscillation $$u = Rcos(omega_0t - delta), qquad R = sqrt{c_1^2 + c_2^2}, qquad andelta = rac{c_2}{c_1}.$$ Period $T = 2pi/omega_0$. The amplitude never decays.

Damped motion — three cases

Everything is decided by $gamma^2$ against $4mk$:

  • $gamma^2 > 4mk$ — overdamped. Two negative real roots, no oscillation.
  • $gamma^2 = 4mk$ — critically damped. Repeated root, no oscillation, fastest return.
  • $gamma^2 < 4mk$ — underdamped. Complex roots; oscillation inside the envelope $e^{-gamma t/2m}$ at quasi-frequency $mu = dfrac{sqrt{4mk - gamma^2}}{2m}$.

In all three cases $u o 0$. Damping decides only whether it oscillates on the way.

Definitions and theorems

Spring constant
A mass $m$ stretching a spring by $L$ at rest gives $k = mg/L$, hence $omega_0 = sqrt{g/L}$.
Amplitude and phase
$c_1cosomega_0t + c_2sinomega_0t = Rcos(omega_0t - delta)$ with $R = sqrt{c_1^2+c_2^2}$ and $ andelta = c_2/c_1$, taking $delta$ in the quadrant matching the signs of $c_1$ and $c_2$.
Damping classification
Overdamped if $gamma^2 > 4mk$, critically damped if $gamma^2 = 4mk$, underdamped if $gamma^2 < 4mk$. Critical damping is $gamma = 2sqrt{mk}$.

Worked example

A 2 kg mass stretches a spring 0.098 m. It is pulled down 0.05 m and released from rest. Find the motion, period and amplitude.

  1. $k = dfrac{mg}{L} = dfrac{2(9.8)}{0.098} = 200$ N/m.
  2. $2u'' + 200u = 0 Rightarrow u'' + 100u = 0$, so $omega_0 = 10$ rad/s.
  3. $u = c_1cos 10t + c_2sin 10t$. Measuring $u$ downward positive: $u(0) = 0.05$ gives $c_1 = 0.05$; $u'(0) = 0$ gives $c_2 = 0$.
  4. $u = 0.05cos(10t)$ m.
  5. Amplitude $R = 0.05$ m, period $T = 2pi/10 approx 0.628$ s. Released from rest at maximum displacement, so the phase is zero.

Common mistakes

  1. Mixing unit systems. $g = 9.8$ m/s² with metres and kilograms, or $g = 32$ ft/s² with feet and slugs. A stretch given in centimetres or inches must be converted first.
  2. Using the weight as the mass. In pounds, weight $= mg$, so $m = W/32$ slugs. In SI a mass given in kilograms is already the mass.
  3. Computing $R$ from the initial conditions directly. $R = sqrt{c_1^2 + c_2^2}$ uses the coefficients; $c_2 = u'(0)/omega_0$, not $u'(0)$.
  4. Getting the phase in the wrong quadrant. $arctan$ returns values in $(-pi/2, pi/2)$; check the signs of $c_1$ and $c_2$ and add $pi$ when needed.
  5. Calling a damped system's oscillation frequency $omega_0$. Damping lowers it to the quasi-frequency $mu < omega_0$.

Practice it

Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.

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