SAME IMPULSE
Same push integrated. Different motion.
One mass, spring, and damper. Three ways to deliver the same applied impulse. Move the second pulse: its timing can reinforce or quiet the motion already under way.
The damped period is 2.002 s. Compare a half-cycle with a whole cycle.
Watch when the push arrives.
Same scales · time 0–8 s · positive motion is to the right
Scrub time to inspect · plots show the complete histories
Shaded force area = applied impulse. The vertical cursor links force, motion, and displacement. The dotted line marks 3 s, when every applied force has ended.
The second push meets a moving mass.
All applied forces point right, yet the spring can pull the mass left. Equal applied impulse fixes the force–time area; it does not fix the displacement history, peak displacement, or remaining vibration.
One oscillator, unchanged
- Mass m
- 1 kg
- Stiffness k
- π² N/m
- Damping c
- 0.3 N·s/m
m x″ + c x′ + k x = F(t)
x(0) = 0, x′(0) = 0
Read the study
Same area, different timing
A delivers one smooth pulse over 0.2–0.8 s. B spreads its push over 0–3 s. C splits the impulse equally between two 0.6 s pulses; its first center is at 0.5 s and the second follows by the selected spacing. Every force is zero outside the common 0–3 s window.
A pulse starting at a, with width w and impulse j, is F(t) = (j/w)[1 − cos(2π(t − a)/w)] on its support and zero elsewhere. Its exact area is j. A and B each use j = 1 N·s; C uses j = 0.5 N·s twice.
Applied impulse is not net impulse
The displayed 1 N·s counts the applied force F alone. Net force also includes −kx and −cx′. The momentum change is the integral of that net force, so equal applied impulse does not force equal final velocity.
Work is another integral, ∫F dx = ∫F x′ dt. Because the mass moves differently while each force acts, equal applied impulse also does not imply equal work or equal energy supplied.
How the numbers are computed
Each pulse is solved analytically as a constant-plus-cosine particular solution with a homogeneous term enforcing zero starting displacement and velocity. After the pulse ends, its response follows the exact free-oscillation solution. Linear superposition adds C’s two responses. Animation only chooses the time at which that solution is displayed.
Peak |x| is the largest displacement magnitude over the displayed 0–8 s interval. Velocity sign changes bracket extrema, which are refined by bisection. Curves are sampled for drawing only; changing playback speed does not change the solution.
What “residual at 3 s” means
Once the force ends, x(t) = e−α(t−3)[x₃ cos(ωd(t−3)) + ((v₃ + αx₃)/ωd) sin(ωd(t−3))], where α = c/(2m) and ωd = √(k/m − α²).
The reported residual is the envelope amplitude √(x₃² + ((v₃ + αx₃)/ωd)²) at 3 s. It is neither the instantaneous displacement nor a claim that this envelope is reached. It subsequently decays at the same rate in every case.
Changing C’s spacing changes both phase and decay time before this common measurement. A near half-cycle spacing leaves less vibration than a near whole-cycle spacing in this construction; damping prevents perfect cancellation.