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:

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?

Hint:

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

$E(t)=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2=\frac{1}{2}mv\mathrm{\_}{max}^2=\frac{1}{2}k(A(t){)}^2$

since the potential energy of a spring is given by

$U_s=\frac{1}{2}k{x}^2$.

Attribution#

Problem is licensed under the CC-BY-NC-SA 4.0 license.