Visualize and explore how damping, mass, and stiffness shape system dynamics
The equation of motion for a mass attached to a spring with damping is:
where:
The total solution splits into two complementary parts:
yc (Complementary/Natural Response): How the system vibrates on its own, determined solely by m, c, and k. This is the solution to the homogeneous equation my'' + cy' + ky = 0.
yp (Particular/Forced Response): How the system responds to the external input F(t). This depends on the forcing function.
Total Solution: y(t) = yc(t) + yp(t)
Watch the physical system move in real time. Adjust parameters, then press Play to see the mass oscillate, with force arrows and a live displacement trace.
The system's response is determined by the discriminant D = c² − 4mk:
Characteristic Equation: mr² + cr + k = 0
The system oscillates with decreasing amplitude. Like a guitar string after being plucked.
Natural Frequency: ωn = 2.0 rad/s
Damping Ratio: ζ = 0.25
Step 1: Write the equation of motion.
$$mx'' + cx' + kx = 0 \quad \Longrightarrow \quad x'' + 4x' + 3x = 0$$Step 2: Form and solve the characteristic equation.
$$r^2 + 4r + 3 = (r+1)(r+3) = 0 \quad \Longrightarrow \quad r_1 = -1,\ r_2 = -3$$Step 3: Classify. The discriminant is $c^2 - 4mk = 16 - 12 = 4 > 0$, so the system is overdamped: two distinct negative real roots, no oscillation.
Step 4: General solution.
$$x(t) = c_1 e^{-t} + c_2 e^{-3t}$$Step 5: Apply the initial conditions.
$$x(0) = c_1 + c_2 = 1$$ $$x'(t) = -c_1 e^{-t} - 3c_2 e^{-3t} \quad \Longrightarrow \quad x'(0) = -c_1 - 3c_2 = 0$$From the second equation $c_1 = -3c_2$; substituting into the first gives $-3c_2 + c_2 = 1$, so $c_2 = -\tfrac{1}{2}$ and $c_1 = \tfrac{3}{2}$.
Final Solution:
$$\boxed{x(t) = \frac{3}{2}e^{-t} - \frac{1}{2}e^{-3t}}$$Step 1: Equation of motion and characteristic equation.
$$x'' + 4x' + 20x = 0 \quad \Longrightarrow \quad r^2 + 4r + 20 = 0$$Step 2: Solve for the roots.
$$r = \frac{-4 \pm \sqrt{16 - 80}}{2} = \frac{-4 \pm \sqrt{-64}}{2} = -2 \pm 4i$$Step 3: Classify. Since $c^2 - 4mk = 16 - 80 = -64 < 0$, the system is underdamped. The decay rate is $\alpha = 2$ and the quasi-frequency is $\mu = 4$ rad/s.
Step 4: General solution.
$$x(t) = e^{-2t}\big(c_1\cos(4t) + c_2\sin(4t)\big)$$Step 5: Apply $x(0) = 0$.
$$x(0) = c_1 = 0 \quad \Longrightarrow \quad x(t) = c_2 e^{-2t}\sin(4t)$$Step 6: Apply $x'(0) = 8$.
$$x'(t) = c_2\big(-2e^{-2t}\sin(4t) + 4e^{-2t}\cos(4t)\big) \quad \Longrightarrow \quad x'(0) = 4c_2 = 8$$ $$c_2 = 2$$Final Solution:
$$\boxed{x(t) = 2e^{-2t}\sin(4t)}$$Step 1: Characteristic equation.
$$x'' + 4x' + 4x = 0 \quad \Longrightarrow \quad r^2 + 4r + 4 = (r+2)^2 = 0$$Step 2: Repeated root. $r = -2$ (double), because $c^2 - 4mk = 16 - 16 = 0$. This is exactly critical damping: $c = c_{cr} = 2\sqrt{mk} = 4$.
Step 3: General solution for a repeated root.
$$x(t) = (c_1 + c_2 t)e^{-2t}$$Step 4: Apply the initial conditions.
$$x(0) = c_1 = 2$$ $$x'(t) = c_2 e^{-2t} - 2(c_1 + c_2 t)e^{-2t} \quad \Longrightarrow \quad x'(0) = c_2 - 2c_1 = 0$$ $$c_2 = 2c_1 = 4$$Final Solution:
$$\boxed{x(t) = (2 + 4t)e^{-2t}}$$Step 1: Find the complementary solution.
$$r^2 + 4 = 0 \quad \Longrightarrow \quad r = \pm 2i \quad \Longrightarrow \quad x_c(t) = c_1\cos(2t) + c_2\sin(2t)$$The natural frequency is $\omega_n = 2$ rad/s — and the forcing frequency is also $2$ rad/s. This is pure resonance.
Step 2: Choose the trial form (multiply by $t$). Because $\cos(2t)$ already appears in $x_c$, we must use
$$x_p(t) = t\big(A\cos(2t) + B\sin(2t)\big)$$Step 3: Substitute. Differentiating twice and simplifying gives
$$x_p'' + 4x_p = -4A\sin(2t) + 4B\cos(2t)$$Step 4: Match coefficients with $3\cos(2t)$.
$$-4A = 0,\qquad 4B = 3 \quad \Longrightarrow \quad A = 0,\ B = \frac{3}{4}$$ $$x_p(t) = \frac{3}{4}t\sin(2t)$$Step 5: Apply the initial conditions to $x = x_c + x_p$.
$$x(0) = c_1 = 0$$ $$x'(t) = 2c_2\cos(2t) + \frac{3}{4}\big(\sin(2t) + 2t\cos(2t)\big) \quad \Longrightarrow \quad x'(0) = 2c_2 = 0$$Final Solution:
$$\boxed{x(t) = \frac{3}{4}t\sin(2t)}$$These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Step 1: Form the characteristic equation. For $mx(t)'' + cx'(t) + kx(t) = 0$:
Step 2: Factor and find roots.
Step 3: Classify the damping. Check the discriminant: $D = c^2 - 4mk = 64 - 64 = 0$. Since $D = 0$, this is critically damped.
Step 4: Interpret the classification. The damping coefficient $\zeta = \dfrac{c}{2\sqrt{mk}} = \dfrac{8}{2\sqrt{2 \cdot 8}} = \dfrac{8}{8} = 1$. When $\zeta = 1$, the system is critically damped — it returns to equilibrium as quickly as possible without oscillating.
Step 5: General solution. For a repeated root $r_0 = -2$:
The term $(c_1 + c_2 t)$ decays exponentially as $e^{-2t}$, ensuring the system settles without bouncing.
Step 1: Form the characteristic equation.
Step 2: Use the quadratic formula.
Step 3: Classify the damping. The discriminant is $D = c^2 - 4mk = 4 - 20 = -16 < 0$. This is underdamped ($\zeta = \dfrac{2}{2\sqrt{5}} \approx 0.447 < 1$). The system oscillates with decreasing amplitude.
Step 4: General solution. From $r = -1 \pm 2i$, we have $\alpha = -1$ and $\beta = 2$:
Step 5: Apply initial condition $x(0) = 1$.
Step 6: Find the derivative.
Step 7: Apply initial condition $x'(0) = 0$.
Step 8: Particular solution.
The exponential envelope $e^{-t}$ damps the oscillation at frequency $\beta = 2$ rad/s. As $t \to \infty$, $x(t) \to 0$.
Test your understanding with these multiple-choice questions. Click a choice to see if you're correct.
Correct! $c^2 - 4mk = 36 - 32 = 4 > 0$, so the roots $r = -2, -4$ are real, distinct and negative: overdamped, no oscillation.
Correct! $\omega_n = \sqrt{k/m} = \sqrt{36/4} = \sqrt{9} = 3$ rad/s.
Correct! $2\sqrt{mk} = 2\sqrt{16} = 8$, so $\zeta = 4/8 = 0.5 < 1$: underdamped, the mass oscillates inside a decaying envelope.
Correct! $\omega_n = 3$ equals the forcing frequency, so $\cos(3t)$ and $\sin(3t)$ are already in $x_c$. Multiply the trial form by $t$ once. (Solving gives $x_p = \tfrac{1}{3}t\sin(3t)$.)
Correct! Critical damping requires $c^2 - 4mk = 0$, i.e. $c = 2\sqrt{mk} = 2\sqrt{2 \cdot 50} = 2\sqrt{100} = 20$ N·s/m.
Correct! The quasi-frequency is the imaginary part $\mu = 3$ rad/s, so $T_d = 2\pi/\mu = 2\pi/3 \approx 2.09$ s.
Critically damped suspensions provide the smoothest ride—the car returns to level without bouncing. Too much damping makes it feel stiff; too little makes it bouncy.
Tall buildings act as damped oscillators. Engineers must understand resonance—when the earthquake frequency matches the building's natural frequency, amplification occurs.
A speaker cone is a driven damped oscillator. The forcing is the electrical signal; the response is sound. Proper damping prevents distortion.
A critically damped door closer closes smoothly without slamming or bouncing. Adjust the damping coefficient to control closure speed.
Section 2.1 (Homogeneous DEs): You learned to find yc, the natural response determined by the characteristic equation mr² + cr + k = 0. The three cases (overdamped, critically damped, underdamped) directly correspond to the three cases of the discriminant.
Section 2.2 (Undetermined Coefficients): You learned to find yp for common forcing functions like constants and sinusoids. Use the method based on the form of F(t).
Here: You see both working together in real time. The sliders let you explore how changes in m, c, and k affect the natural response, and how different forcing functions shape the forced response. The visualization makes the theory come alive.