Skip to content
KinPilot™Learn & Explore
Menu

Grades 9–12 · Machines

How can a moving mass make a tall building sway less?

A tall building can sway noticeably in wind even while its structure remains within its designed limits. A tuned mass damper addresses a selected part of this vibration. It adds a secondary mass that can move relative to the building through a restoring mechanism and dampers. Taipei 101’s visible suspended mass is a celebrated example.

M49 visual explanationThe secondary mass has relative displacement 0.2 m and relative velocity 0.1 m/s. The attachment force is −kq−cv; its reaction acts on the building. This snapshot does not predict a whole building’s vibration reduction.Relative motion transfers forces and energyLocal attachment model; not a tower responseBuilding attachmentMass100 kgk: 400 N/mc: 40 N s/mq = 0.2 m; v = 0.1 m/sMass: -84 N; building: 84 NDamping power: 0.4 W; spring stores energy

In the diagramRelative motion transfers forces and energyLocal attachment model; not a tower responseBuilding attachmentMass100 kgk: 400 N/mc: 40 N s/mq = 0.2 m; v = 0.1 m/sMass: -84 N; building: 84 NDamping power: 0.4 W; spring stores energy

The mass moves relative to the building through a spring and damper. The example shows equal and opposite attachment forces and nonnegative damping power, not an assumed motion phase.

The building and added mass form coupled oscillators. Designers choose the mass and restoring stiffness so that the secondary oscillator’s natural frequency lies near a target building mode. Motion can then exchange energy between the building and the secondary oscillator. The response depends on excitation frequency, mass ratio, tuning and damping; it is not described by saying the mass always moves exactly opposite to the tower.

A small local model makes the force transfer explicit. Let q be the mass’s displacement relative to the building, and v its relative velocity. With an ideal spring and viscous damper, the attachment force on the mass is F = −kq − cv. The building receives the opposite attachment force. For the illustration’s k = 400 newtons per meter, c = 40 newton-seconds per meter, q = 0.2 meter and v = 0.1 meter per second, that force is −84 newtons.

The spring can store energy and return it. The viscous damper instead dissipates energy at a rate P = cv², which is 0.4 watt in this small example and never negative. Real systems require a complete dynamic analysis and enough travel for the moving mass. A single force snapshot explains the coupling but cannot predict the building’s eventual vibration reduction.

About this explanation

Starting knowledge: Use Hooke’s law, relative displacement and velocity, frequency and mechanical power.

Sources

  • TAIPEI 101 — Observatory Features: Tuned Mass Damper
  • MIT, Alexander H. Slocum — Precision Machine Design: Topic 11, Damping
← Grades 9–12 topics