Exponential Damping 4#

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:

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

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#

What is the period of oscillation of this spring?


It is useful to remember that:

(13)#\[\begin{equation} \omega = 2\pi f = \frac{2\pi}{T} \end{equation}\]

where \(T\) is the period and \(\omega\) is the angular frequency.

Answer Section#

Please enter in a numeric value in s.



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