Equation of motion of a dampened pendulum

[Physics Dad]

Homework Statement

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.

Homework Equations

F=ma
F=-kx
x(t)=A₀e^αt

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)=A₀e^αt I get...

x(t)=A₀e^(-b/2m)t+e^iϑt

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

Any help gratefully received!

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

F=ma
F=-kx
x(t)=A₀e^αt

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)=A₀e^αt I get...

x(t)=A₀e^(-b/2m)t+e^iϑt

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

Any help gratefully received!

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##).

 

[Physics Dad]

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

 

[Ray Vickson]

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

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.