Damped and Forced Oscillations
Senior 5 · Physics
Damped and Forced Oscillations
Lesson 1 of 3
25 min
Pull a child's swing back and let go. In Unit 1 an ideal oscillator would keep swinging forever with the same amplitude. A real swing does not: each swing is a little smaller than the one before, and after a minute or so it hangs still. Air resistance and friction at the pivot take a little energy from every cycle. Engineers sometimes fight this loss, as in a clock, and sometimes add as much of it as they can, as in the shock absorbers of a car or minibus. This lesson describes the motion of an oscillator when a resistive force acts, and the three quite different ways it can come to rest.
Any force that opposes the motion and removes energy from an oscillator is called a damping force. For a body moving slowly through a fluid (air, water, oil), the damping force is close to proportional to the velocity:
where is the damping constant, in (or ), and the minus sign means the force always opposes the velocity. Adding it to the restoring force of a spring, Newton's second law becomes
It is useful to write , the natural angular frequency the system would have without damping, and , which measures how strong the damping is. The equation then reads .
When the damping is small (), the solution is an oscillation whose amplitude decays exponentially:
You can check this by substituting it into the equation. Two things happen. The amplitude shrinks by the same factor in every equal interval of time, so the ratio of any two successive peaks is constant. The angular frequency is a little smaller than , so the period is a little longer; for light damping the difference is tiny.
Definition
A damped oscillation is one whose amplitude decreases with time because a resistive force removes energy from the system. The damping constant relates the damping force to the velocity, . Example: a mass on a spring hanging in a beaker of oil oscillates a few times with shrinking amplitude and stops; the same mass in air swings for minutes.
Worked example 1: a lightly damped spring
A 0.50 kg mass on a spring of stiffness moves in air with damping constant . Find (a) and , (b) the damped period and compare it with the undamped period, (c) the fraction of the starting amplitude left after 5.0 s.
Solution
(a) ; .
(b) , so . Undamped: . The difference is less than 0.1%.
(c) . About 14% of the amplitude is left after 5 s, about 8 oscillations.
As the damping grows, the motion changes character. The deciding comparison is between and .
| Type | Condition | Behaviour after release | Example |
|---|---|---|---|
| Light (underdamped) | , that is | oscillates about equilibrium with exponentially decreasing amplitude | a pendulum or tuning fork in air |
| Critical | , that is | returns to equilibrium in the shortest possible time without overshooting | the needle of a moving-coil meter; a car's suspension is designed close to this |
| Heavy (overdamped) | creeps back to equilibrium slowly, without oscillating | an automatic door closer; a mass on a spring in thick oil |
The critical damping constant follows from setting . Notice from the graph that heavy damping is not the quickest way to stop an oscillation: too much damping makes the return to equilibrium slower.
Definition
Critical damping is the amount of damping that brings a displaced system back to its equilibrium position in the shortest possible time without any oscillation. For a mass–spring system the critical damping constant is . Example: for and , , 25 times the damping in Worked example 1.
Every car, minibus and motorcycle has springs to absorb bumps in the road and dampers, often called shock absorbers, to stop the body bouncing afterwards. Without dampers the body would bounce up and down for many cycles after each pothole; with far too much damping it would ride the bump stiffly and pass every jolt to the passengers.
Worked example 2: choosing a shock absorber
One corner of a car carries a sprung mass of 300 kg on a spring of stiffness (illustrative values). Find (a) the natural frequency of the bounce, (b) the damping constant for critical damping.
Solution
(a) , so .
(b) . With this damper the body would settle after a bump in about one natural period, without bouncing. Worn shock absorbers lower : the car then bounces several times after each bump, and its tyres can briefly lose grip.
Worked example 3: successive amplitudes
A lightly damped oscillator has successive peaks of 8.0 cm and 6.0 cm on the same side. (a) Predict the next two peaks. (b) After how many more cycles is the amplitude first below 1.0 cm?
Solution
(a) For exponential decay the ratio of successive peaks is constant: . The next peaks are and . (The amplitude does not drop by 2.0 cm each time; that would be linear decay.)
(b) After cycles from the 8.0 cm peak, . Set : . So the first peak below 1.0 cm comes 8 cycles after the 8.0 cm peak (it is 0.80 cm).
Warning
Common mistakes. (1) Thinking the heaviest damping stops a system fastest: critical damping does; heavy damping is slower. (2) Assuming the amplitude falls by equal amounts each cycle: it falls by equal ratios. (3) Saying damping changes the period a lot: for light damping is very close to . (4) Calling any decay "critical": critical damping is a precise value, .
Tip
Compare with . Smaller: it oscillates (light). Equal: fastest return, no overshoot (critical). Larger: slow creep back (heavy).
Critical Damping (xmdemo). Watch the same oscillator with too little, just enough and too much damping, and notice which one settles first.
Practice
Key points
Sources and credits
The three diagrams, the worked examples and the practice exercises are written by Ganzaa.