# 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 weâ€™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)#$$$F\_{net,x} = -b\frac{dx}{dt} - kx=ma = m\frac{d^2x}{dt^2}$$$

The solution of this differential equation is found to be $$x(t) = Ae^{-\frac{bt}{2m}} \cos(\omega t) = Ae^{-\frac{t}{2\tau}} \cos(\omega t)= A(t) \cos(\omega t)$$ , 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:

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

since the potential energy of a spring is given by

$$U_s = \frac{1}{2} kx^2$$.