Logarithmic decrement of a lightly damped oscillatorby kraigandrews Tags: damped, decrement, lightly, logarithmic, oscillator 

#1
Jun2711, 10:29 PM

P: 108

1. The problem statement, all variables and given/known data
The logarithmic decrement δ of a lightly damped oscillator is defined to be the natural logarithm of the ratio of successive maximum displacements (in the same direction) of a free damped oscillator. That is, δ = ln(An/An+1) where An is the maximum displacement of the nth cycle. Derive the simple relationship between δ and Q. Find the spring constant k and damping constant b of a damped oscillator with mass m, frequency of oscillation f and logarithmic decrement δ. [Data: m = 4.0 kg; f = 0.9 Hz; δ = 0.029.] First, the spring constant k... Also, the damping constant b... 2. Relevant equations [itex]\beta[/itex]=b/(2m) Q=[itex]\omega_{o}/(2\beta) 3. The attempt at a solution Given the diff eq: d^{2}x/dt^{2}+2[itex]\beta[/itex](dx/dt)+[itex]\omega[/itex]_{o}^{2}x=0 I can solve this to find x(t), however I feel this is irrelevant because no initial condition or boundary conditions are given, so I am kinda lost here as to where go or to start at for that matter. Any suggestions are greatly appreciated, Thanks 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Jun2811, 02:08 AM

HW Helper
P: 2,324

The initial conditions don't matter. Can you write the exponential term of the equation of motion in terms of Q? This term controls the amplitude of the oscillator, so the ratio between successive amplitudes is just the ratio between successive exponential terms.



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