Hi,(adsbygoogle = window.adsbygoogle || []).push({});

For an undamped mass, spring system subject to a harmonic force, the equation is: ma+kx=Fsin(ωt)

where a=d^2/dx^2

After solving the diff eqns, the steady state amplitude is:

X=(F/m)/(ω^2-ω0^2)

where ω is the frequency and ω0 is the natural frequency =sqrt(k/m)

according to the amplitude equation, the amplitude will increase if you increase the stiffness.

I am struggling to understand why this is true, as i understand it, the spring should be resisting the motion of the mass and hence the stiffer the spring the less it moves. Think about if you increase the stiffness to infinity, equivelant to placing the mass on a rigid floor, the displacement of the mass should surely reduce to zero. Am I missing something?

Anyone able to explain?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Spring Stiffness and Amplitude, not understanding

Loading...

Similar Threads - Spring Stiffness Amplitude | Date |
---|---|

Why are stiff springs more responsive? | Feb 1, 2016 |

How to find beam spring stiffness coefficient? | Sep 13, 2015 |

Rotary system stiffness/travel question | Jun 15, 2015 |

Ansys workbench-spring with variable stiffness | Feb 12, 2014 |

Determine the Spring Design given Spring Stiffness of 400 N/m | May 1, 2013 |

**Physics Forums - The Fusion of Science and Community**