How to Solve for x(t)/x0 in a Damped Oscillator with Initial Values?

Click For Summary
The discussion revolves around solving the equation for a damped oscillator, specifically d2x/dt2 + 2βdx/dt + ω0^2 x = 0, with given parameters ω0 = 1.0 s−1 and β = 0.54 s−1. The initial conditions are x(0) = x0 and v(0) = 0, and the goal is to determine x(t)/x0 at t = 2π/ω0. The solution involves defining μ as an imaginary number, indicating an underdamped system, which complicates the calculation of coefficients A1 and A2. By applying Euler's formula, the imaginary components can be managed, leading to a real solution for x(t).
kraigandrews
Messages
105
Reaction score
0

Homework Statement



The equation for a damped oscillator is d2x/dt2+2βdx/dt +ω02 x = 0. Let ω0=1.0 s−1 and β = 0.54 s−1. The initial values are x(0) = x0 and v(0)=0.
Determine x(t)/x0 at t = 2π/ω0.


Homework Equations



the solution to equation is given by;

x(t)=e-\betat(A1et\mu+A2e-t\mu)

where \mu=\sqrt{\beta<sup>2</sup>-\omega<sub>o</sub><sup>2</sup>}

The Attempt at a Solution



A1=1/2(xo+(xo\beta)/\mu)
A2=1/2(xo-(xo\beta)/\mu)

The problem I am running into is that the parameter I defined as \mu is imaginary for this case, which keeps throwing me off. My only guess is to ignore the term multiplied by A1 because it is not real, then use only the A2 term and its multiplier because of the -t in its exponent making -i =1. I do not know if this correct and also even the constants A1 and A2 have an i in them as wel.
 
Physics news on Phys.org
The fact that μ is imaginary tells you you have an underdamped system.

Apply the initial conditions to solve for A1 and A2. The coefficients will be complex values. Then use Euler's formula, e^{i\theta} = \cos \theta + i \sin \theta, to express the solution in terms of sines and cosines. You'll find everything works out so the i's cancel and x(t) is real.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
9
Views
2K
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
815
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
Replies
7
Views
2K