Free Mechanical Vibrations
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.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
A mass of 1 kg hangs at rest on a spring and stretches it 0.4 m. Take $g = 9.8\ \text{m/s}^2$ and neglect damping.
Solution
- At rest the spring force balances gravity: $kL = mg$, so $k = \dfrac{mg}{L} = \dfrac{1(9.8)}{0.4} = 24.5$ N/m.
- The equation of motion is $mu'' + ku = 0$, i.e. $u'' + \dfrac{k}{m}u = 0$.
- $\omega_0 = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{g}{L}} = \sqrt{\dfrac{9.8}{0.4}} = 4.9497$ rad/s.
- Notice the mass cancels — for a spring stretched by its own hanging weight, the frequency depends only on the stretch.
- Period $T = \dfrac{2\pi}{\omega_0} = 1.2694$ s.
Next problem, fresh numbers → No account needed. Your accuracy is tracked on this device.
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: $$kL = mg \quad\Rightarrow\quad k = \frac{mg}{L}.$$ 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 $u = c_1\cos\omega_0t + c_2\sin\omega_0t$, which is the single oscillation $$u = R\cos(\omega_0t - \delta), \qquad R = \sqrt{c_1^2 + c_2^2}, \qquad \tan\delta = \frac{c_2}{c_1}.$$ Period $T = 2\pi/\omega_0$. 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 $u \to 0$. Damping decides only whether it oscillates on the way.
Definitions and theorems
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.
- $k = \dfrac{mg}{L} = \dfrac{2(9.8)}{0.098} = 200$ N/m.
- $2u'' + 200u = 0 \Rightarrow u'' + 100u = 0$, so $\omega_0 = 10$ rad/s.
- $u = c_1\cos 10t + c_2\sin 10t$. Measuring $u$ downward positive: $u(0) = 0.05$ gives $c_1 = 0.05$; $u'(0) = 0$ gives $c_2 = 0$.
- $u = 0.05\cos(10t)$ m.
- Amplitude $R = 0.05$ m, period $T = 2\pi/10 \approx 0.628$ s. Released from rest at maximum displacement, so the phase is zero.
Common mistakes
- 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.
- 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.
- 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 $u'(0)$.
- 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.
- 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.