Does logarithmic decrement depend on initial amplitude?by guest1234 Tags: amplitude, decrement, depend, initial, logarithmic 

#1
Nov2412, 07:16 PM

P: 24

Lets say we have a weight attached to a spring. When releasing it under water (whole motion occours under water) at different initial amplitudes, will logarithmic decrement be the same? And will the period change?
I think it should be the same for different initial amplitudes. But [itex]\Theta=\frac{T_1}{t}\ln\frac{A_1}{A(t)}[/itex], where [itex]T_1[/itex] is period of first cycle, [itex]t[/itex] is time until amplitude is [itex]A(t)[/itex] and [itex]A_1[/itex] is initial amplitude, suggests that when initial amplitude increases, logarithmic decrement must increase as well, or the ratio [itex]\frac{T_1}{t}[/itex] must decrease. So... which one is correct? 



#2
Nov2412, 08:01 PM

Engineering
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P: 6,341

Your formula for log dec is only correct when the time interval is an exact multiple of the oscillation period.
In real life, you can use the formula approxiamately for any time interval, by drawing a smooth curve through the maximum amplitude of each cycle, and using that curve for the amplitudes "A" in the formula. Also, the log dec is only constant if the damping force is proportional to velocity. That is the "standard" equation that you study for single degree of freedom damped systems, but in real life damping forces are often NOT proportional to velocity. For example the damping force caused by viscosity of a fluid (like your water example) is approximately proportional to velocity squared, except at very low velocities. As another example, a frictional damping force that obeys Coulomb's law of friction is constant (indepdendent of velocity). For friction damping, the motion will stop completely after a finite number of oscillations, which can't happen with the log dec is constant. 


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