# Thread: Damping ratio

1. ## Damping ratio

Hi guys,
Revision for an exam tomorrow, I hope someone can help?

I've been given the answers but I can't work out how to find the damping ratio without being given a velocity or distance for the mass to travel?

A mass of 6kg is suspended on a spring and set oscillating. it is observed that amplitude reduces to 5% of its initial value after 2 oscillations, which takes 0.57 seconds.

Determine the following.

The damping ratio (0.43)
The natural frequency (3.51)
The actual frequency (3.17)
The spring stiffness (2915N/m)
The critical damping coefficient (264.5Ns/m)
The actual damping coefficient (113.7Ns/m)

I can use the damping ratio to help solve some of the other values once I know how to work it out.

Any help appreciated

2. ## Re: Damping ratio

OK this is what I have,
Natural frequency = 2/0.57 = 3.51Hz
Spring stiffness 3.51Hz = 1/2pi x square root of K/m (k works out to be about 2915)
Critical damping square root of 4mk = 264.5

I need the damping ratio to find actual damping and actual frequency, I don't think I can do it without a velocity?
The 5% of amplitude must come into it somehow?