# Homework Help: Equation of motion of a dampened pendulum

1. May 8, 2017

1. The problem statement, all variables and given/known data

Hi, I have the following question...

A pendulum consisting of a mass of 1kg is suspended from a string of length L. Air resistance causes a damping force of bv where b = 10-3 N/m

1) Derive and solve the equation of motion
2) Calculate the fractional decrease in amplitude of the pendulum oscillations if the pendulum is operated for 2 hours.

2. Relevant equations

F=ma
F=-kx
x(t)=A0eαt

3. The attempt at a solution

I think I have the derivation handled...

F=ma=-kx-bv
ma+bv+kx=0
m(d2x/dt2)+b(dx/dt)+kx=0
d2x/dt2+(b/m)(dx/dt)+(k/m)x=0

As this is a pendulum, I know x=Lsinθ and for small θ sinθ≈θ so...

d2θ/dt2+(b/m)(dθ/dt)+(k/m)θ=0

I also know that k=mg/L so...

d2θ/dt2+(b/m)(dθ/dt)+(g/L)θ=0

Therefore, I get the equation of motion for the dampened pendulum to be:

d2θ/dt2+(b/m)(dθ/dt)+(g/L)θ=0

Now to solve the equation, I can turn it into a quadratic:

α=(-(b/m)±(√(b2/m2)-(4g/L))/2

I can tidy this up a little, so...

α=-(b/2m)±(√(b2/2m2)-(4g/L))

I know that √g/L = ω so and -√g/L = iω which I can take out as a factor so and tidy up the fraction inside the radical, so...

α=-(b/2m)±iω√1-(b2L/2)

I also know that I can let ϑ=ω√1-(b2/4m2ω2) so...

α=-(b/2m)±iϑ

I am assuming that this is the equation solved as I don't have a value for L and so can't go any further (can I?)

As for calculating the fractional decrease, if I sub this back into the equation:

x(t)=A0eαt I get...

x(t)=A0e(-b/2m)t+eiϑt

But I really don't know where to go from here, if, if anywhere?

Do I simply rearrange the first part so....

x(t)/A0=e-18/5 so...

x(t)/A0=0.0273

2. May 8, 2017

### Ray Vickson

You can write the solution either as
$$x = A_1 e^{-rt}e^{i st} + A_2 e^{-rt} e^{-ist},$$
or as
$$x = B_1 e^{-rt} \cos(st) + B_2 e^{-rt} \sin(st),$$
where $r$ and $s$ are related to your constants $m, k, b$. If you choose the second form you just need to determine the constants $B_1, B_2$, which you can do from the initial conditions (position and velocity at $t = 0$).

Last edited: May 9, 2017
3. May 9, 2017

thanks for that,

I kind of get what you're saying, using the second equation, when t=0, x=B1

I am still confused as to what to do when t=7200 though

4. May 9, 2017

### Ray Vickson

You need to use one more piece of information in order to determine the constant $B_2$. For example, if the pendulum bob is released at zero velocity from initial position $x_0$ then you have $x(0) = x_0$ and $\dot{x}(0) = 0$. From those two conditions you can get both $B_1$ and $B_2$ in terms of $x_0, r,s$.

Then, you just need to plug in $t = 7200$ into your formula for $x(t)$ and evaluate it.

Last edited: May 9, 2017