Exponential Damping 3#

A \({{params_m}}\) kg mass oscillates on a \({{params_k}}\) N/m spring. The damping constant of this spring is \(b\) = \({{params_b}}\) kg/s.

Useful Info#

For slowly moving objects we’ve seen that the drag force grows in proportion to the velocity, \(\overrightarrow{D} = -b\overrightarrow{v}\), where \(b\) is the damping constant and \(\overrightarrow{v}\) is the velocity of the object.

The net force acting on a slowly moving mass attached to a massless spring in the presence of a drag force (for motion along \(x\) relative to an equilibrium point \(x_0\)) can be written as:

(11)#\[\begin{equation} F\_{net,x} = -b\frac{dx}{dt} - kx=ma = m\frac{d^2x}{dt^2} \end{equation}\]

The solution of this differential equation is found to be \begin{equation} x(t) = Ae^{-\frac{bt}{2m}} \cos(\omega t) = Ae^{-\frac{t}{2\tau}} \cos(\omega t)= A(t) \cos(\omega t) \end{equation} , where \(A\) is the initial amplitude of the oscillation, \(\tau\) is the time constant, \(A(t)\) is the time-dependent amplitude of the oscillation, and \(\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}\) is the angular frequency of the damped oscillation.

Part 1#

How long does it take for the energy stored in the spring system to reach half of its initial value?


The total energy of the oscillating spring system as a function of time is given by:

\(E(t) = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} mv\_{max}^2 = \frac{1}{2} k(A(t))^2\)

since the potential energy of a spring is given by

\(U_s = \frac{1}{2} kx^2\).

Answer Section#

Please enter in a numeric value in s.



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