Does logarithmic decrement depend on initial amplitude?

AI Thread Summary
Logarithmic decrement does not remain constant across different initial amplitudes when a weight is attached to a spring and released underwater. The relationship between amplitude and logarithmic decrement is influenced by the damping force, which may not be proportional to velocity in real-world scenarios. For instance, damping forces caused by fluid viscosity can vary with the square of the velocity, while frictional damping can be constant. The formula for logarithmic decrement is most accurate when the time interval aligns with the oscillation period. Therefore, the behavior of logarithmic decrement and the period of oscillation can change based on the specific damping conditions present.
guest1234
Messages
38
Reaction score
1
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 \Theta=\frac{T_1}{t}\ln\frac{A_1}{A(t)}, where T_1 is period of first cycle, t is time until amplitude is A(t) and A_1 is initial amplitude, suggests that when initial amplitude increases, logarithmic decrement must increase as well, or the ratio \frac{T_1}{t} must decrease.

So... which one is correct?
 
Physics news on Phys.org
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.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top