Compute the natural frequency of a spring-mass system and explain resonance and isolation.
Why it matters
Anything with mass and stiffness can vibrate: a pump on its mounts, a footbridge under walkers, a car on its suspension, a building in an earthquake. Each has a natural frequency at which it prefers to oscillate. If something pushes it at or near that frequency, the motion can grow far larger than the push alone would cause. This is resonance, and avoiding it (or controlling it) is a routine part of mechanical and structural engineering.
The spring-mass model
The simplest vibrating system is a mass m on a spring of stiffness k (N/m), with no damping. Pull the mass down and let go: the spring force −k·x pulls it back, it overshoots, and it oscillates. Newton's second law gives
m·ẍ + k·x = 0
whose solution is a sine wave with natural circular frequency
ωn = √(k / m) (rad/s)
fn = ωn / (2π) (Hz, cycles per second)
T = 1 / fn (s, period)
Stiffer springs raise the frequency; heavier masses lower it. Doubling the mass lowers fn by a factor of √2, not 2.
Worked example: a machine on springs
A 20 kg compressor sits on four mounts with a combined stiffness of 80,000 N/m.
ωn = √(80,000 / 20) = √4,000 = 63.2 rad/s
fn = 63.2 / (2π) = 10.1 Hz
If the compressor's motor runs at 600 rpm, its forcing frequency is 600/60 = 10 Hz, right on top of the natural frequency. The mounts would amplify the shaking instead of isolating it. Choosing softer mounts lowers fn well below the running speed.
A shortcut from static deflection
Under its own weight the mass compresses the spring by δst = m·g/k. Substituting k = m·g/δst:
fn = (1 / 2π) · √(g / δst)
For the compressor, δst = 20 × 9.81 / 80,000 = 2.45 mm, and √(9.81 / 0.00245) / 2π gives the same 10.1 Hz. This is handy in practice: measure how far something sags under its own weight and you can estimate its natural frequency.
Springs in combination
- Parallel springs (side by side, sharing the load) add: k = k1 + k2.
- Series springs (one on top of another) combine like resistors in parallel: 1/k = 1/k1 + 1/k2.
Four 20,000 N/m mounts under one machine act in parallel: 80,000 N/m. Two 20,000 N/m springs stacked in series give only 10,000 N/m.
Resonance and the frequency ratio
When a harmonic force at frequency f drives the system, the response depends on the frequency ratio r = f / fn:
- r much less than 1: the system follows the force almost statically.
- r near 1: resonance. Without damping the amplitude grows without limit; with damping it peaks at a large finite value.
- r well above 1 (in practice above about √2 ≈ 1.41): the mass barely moves, and less force passes to the floor. This is how vibration isolation works.
Damping
Real systems lose energy to friction, air and material hysteresis, so free vibrations die away. Damping is described by the damping ratio ζ. Many structures have ζ of only a few percent, which is why resonance can still be dramatic: in 2000 the London Millennium Bridge swayed noticeably under crowds of walkers because of a lateral resonance, and it was fixed by adding dampers.
Beams and structures
A beam is not a single spring and mass; it has many natural frequencies (modes). For a first estimate, the stiffness of a beam at a point comes from its deflection formula: a simply supported beam with a central load has k = 48·E·I / L³. Floors are often checked to keep their first natural frequency above a few hertz so that footsteps do not excite them. Real limits and methods come from design guides and codes.
Common mistakes
- Mixing rad/s and Hz. ωn = 2π·fn. Most specifications and machine speeds are in Hz or rpm.
- Using grams or kN without converting. Use kg, N/m and m to get rad/s.
- Adding series springs directly. Series stiffness is always less than the softest spring.
- Assuming damping fixes everything. Light damping still allows large resonant motion; move the natural frequency away from the forcing frequency first.
Recap
- ωn = √(k/m), fn = ωn/2π.
- fn = (1/2π)·√(g/δst) from the static deflection.
- Parallel springs add; series springs combine reciprocally.
- Keep forcing frequencies away from natural frequencies; isolation needs r above about 1.41.
Try it
On the Do tab, finish a natural frequency function and check the compressor example.