Free vibration, natural frequencies and damping

Build a mass-spring-damper idealisation

Choose displacement x measured from equilibrium, velocity v and acceleration a in the same direction. A linear spring supplies restoring force−kx and a viscous damper supplies−cv. With no external drive, the equation is ma+cv+kx=0. The terms have force units: k is in N/m, c in N·s/m, and m in kg.

One displacement coordinate makes this a single-degree-of-freedom model. An actual flexible object may bend in several shapes, each with different motion relationships and frequencies. A single mass-spring representation is useful when one mode dominates in the range of interest; it is not a claim that every point moves identically.

Natural frequency depends on the chosen model

Without damping, the angular natural frequency is ωn=√(k/m). Ordinary frequency fn=ωn/(2π) is in cycles per second, or Hz; period is 1/fn. A larger stiffness raises the frequency and a larger mass lowers it when other model features are held fixed. Initial displacement and velocity select amplitude and phase rather than changing this linear model’s frequency.

The undamped response exchanges spring potential energy and kinetic energy. At a displacement extreme, velocity is zero; near equilibrium, speed is largest. Mechanical energy remains constant only when no damping or other transfer is present. A system can be at equilibrium position while moving rapidly, so position alone does not describe its dynamic state.

Read the time law as a motion pattern

The cosine and sine functions describe repeating patterns. One way to define them is to follow a point counterclockwise around a circle of radius one, measuring θ from the positive horizontal axis: its horizontal coordinate is cosθ and its vertical coordinate is sinθ. Each lies between −1 and +1. The angle θ is in radians; a full turn is 2π radians. You do not need calculus to read the repeating values shown here.

For the undamped model x=A cos(ωnt), A is the positive displacement amplitude, measured from equilibrium to a maximum. The product ωnt is an angle: rad/s multiplied by seconds gives radians. At t=0, cos0=1, so x=A. After a quarter period, the angle is π/2 and x=0. After half a period it is π and x=−A; after a full period it is 2π and x=A again. This is why the period is 2π/ωn, while the displacement amplitude is a separate quantity.

Phase means the position within a repeating cycle when the clock starts. Compare x=10 cos(10t) mm with x=10 sin(10t) mm. Both have amplitude 10 mm and period 2π/10≈0.628 s. The cosine case begins at +10 mm at rest; the sine case begins at equilibrium moving in the positive direction. A sine trace can also be written as a cosine trace shifted by a quarter cycle. These are different initial states of the same undamped model, not evidence of a different stiffness or natural frequency.

For the cosine case, the first equilibrium crossing occurs at about 0.157 s, the negative extreme at 0.314 s and the return to the positive extreme at 0.628 s. At equilibrium its spring energy is zero but its speed is largest. The figure lets you locate those events before evaluating another point in a formula.

Damping can remove oscillation without adding stiffness

Define damping ratio ζ=c/(2√(km)). For 0<ζ<1, the ideal free response oscillates while its envelope decays. Its angular oscillation frequency is ωd=ωn√(1−ζ²). At critical damping ζ=1, the ideal response returns without repeated oscillation; larger damping also avoids oscillation but is not automatically faster for every initial state and response measure.

Viscous damping is an assumed force law, not a description of every frictional or material loss. The plotted comparison uses the same initial displacement and zero initial velocity for both cases. That shared release is essential: changing initial conditions as well as damping would make a simple visual comparison misleading.

The exp notation, also written with the number e≈2.718 as a base, describes a smooth exponential multiplier. For example, exp(−t/τ) starts at 1, becomes about 0.368 after one time constant τ and about 0.135 after two. Its exponent is dimensionless. In the worked damped response, e^(−t) means exp[−(1 s⁻¹)t] when t is expressed in seconds. This factor shrinks the oscillatory motion over time; it does not change the spring stiffness. The bracketed sine/cosine combination in that response also matters: it was chosen to give zero initial velocity, rather than assuming a decaying cosine alone has the same release.

Compare one persistent and one decaying response

For m=1 kg and k=100 N/m,ωn=10 rad/s, fn≈1.59 Hz and undamped period≈0.628 s. Starting at 10 mm displacement with zero velocity, the undamped trace is x=10cos(10t) mm. Its initial spring energy is½×100×0.010²=0.005 J.

With c=2 N·s/m,ζ=0.10 and ωd=√99≈9.95 rad/s. The dashed trace uses x=10e^(−t)[cos(√99t)+sin(√99t)/√99] mm, which preserves that same release condition. After one damped period, about 0.631 s, the next positive peak is about 5.32 mm. The decrease represents energy transferred out of this mechanical oscillation; it is not a decrease of the supplied spring stiffness.

Try the reasoning

1. Does a heavier mass raise ωn if k stays fixed?

   Show the answer and reasoning

   No. Since ωn=√(k/m), the frequency decreases.

2. Why compare identical initial displacement and velocity?

   Show the answer and reasoning

   Otherwise different release conditions could be mistaken for a damping effect.

3. Does one observed frequency establish that a structure has only one mode?

   Show the answer and reasoning

   No. Other modes may be weakly excited, poorly observed or outside the measured range.

4. For the same undamped mass and spring, release from −10 mm at rest instead of +10 mm. Does its frequency change, and how does its trace compare?

   Show the answer and reasoning

   The natural frequency and period remain the same. The new trace is the negative of the old cosine trace: it starts at the opposite extreme, corresponding to a half-cycle phase difference. Changing the initial state does not change m or k.

5. Keep m=1 kg but increase the ideal spring stiffness from 100 to 400 N/m. What happens to the period and the first equilibrium-crossing time after a release from rest?

   Show the answer and reasoning

   ωn=√(400/1)=20 rad/s, twice the old value. The period halves to 2π/20≈0.314 s, and the first equilibrium crossing is one quarter of that, about 0.0785 s. Changing stiffness changes frequency; simply changing release amplitude would not in this linear model.

What to carry forward

Natural frequency, damping and initial state describe different features of free motion.