Vibrations question

Question

Suppressive vibrations (oscillations) happen so that the amplitude of the third vibration is 11,02 times greater than the amplitude of the sixth vibration. The frequency of the vibrations is 0,5 Hz. Find the suppression coefficient β and time tn, during which the amplitude of the vibrations decreases e^n times. Here e is the base of the natural log function.

Answer 1

I am not sure of the notation used in this problem. However, I think what is being referenced is damped harmonic oscillation.

The period is 0.5 Hz which means that the period is 2 s

The third and sixth 'vibrations' are thus separated in time by 6 s.

The amplitude ratio is thus exp( 6 β ) which is equal to 11.02

Solving tor the extinction coefficient β gives β = .4 /s

I think tn is just 1 / β = 2.5 s