A 40 kg child bouncing on a trampoline can be roughly modelled as a damped harmonic oscillator. If the height that the child bounces (the oscillation amplitude) drops to 1/3 its original value in 1.8 s, what is the value of the damping constant?

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The equation describing a damped harmonic oscillator is the following derived from combining Newton's second law (`F=ma=m ddotx`), Hooke's law for springs (`F=-kx`), and the equation governing the force-velocity relation for dampers (`F=-cv = -c dotx`), where c is our damping coefficient (1):

`ddot x + c/m dotx + k/m x = 0`

This equation is a homogeneous secondary differential equation with a certain set of solutions. Given that this is a child on a trampoline and that we know that oscillations follow the input, we can see that the solution will be the standard underdamped solution for this relation, a well known form (1):

`x(t) = A e^( -c/m t/2)cos(omega t + phi)`

The sinusoid term with its frequency and phase in this case are irrelevant, thankfully. We wouldn't be able to solve for them anyway without a known spring constant, k! The important part is the term preceding the sinusoid, where the amplitude is determined. We can use a known initial condition, that at time t=0, the amplitude is A. We also know that at time t=1.8 seconds that the amplitude is A/3. We can use this last relation and the given mass of 40 kg to set up the one that will allow us to solve for the damping constant:

`A/3=Ae^(-c/40 1.8/2)`

Now, we solve for c:

`1/3 = e^(-36c)`

`ln(1/3) = -36c`

`c = ln(3)/36 = 0.03`

So, our damping coefficient c is 0.03 kg/sec.

`1/3 = e^(-c/40*0.9)`

`1/3 = e^(-0.0225c)`

`ln(1/3) = -0.0225c`

`ln(3)/0.0225 = c = 48.8`

The answer is c = 48.8 kg/s.

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