# Spring Differential Equation w/ Damping

1. Mar 31, 2013

### AmagicalFishy

1. The problem statement, all variables and given/known data
A spring with a spring constant of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion where t is time and y(t) is displacement with respect to time.

2. Relevant equations

$$10y'' + 4y' + 20y = 0 \\ y(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \\ e^{i\theta} = cos{\theta} + isin{\theta} \\ y(0) = 1 \\ y'(0) = 0 \\$$

3. The attempt at a solution
The ODE's characteristic equation is:
$$10\lambda ^2 + 4\lambda + 20 = 0 \\ \text{or} \\ 5\lambda ^2 + 2\lambda + 10 = 0$$
Solving for lambda...
$$\lambda = \frac{-1 \pm 7i}{5}$$

Therefore, y(t)—the solution to the ODE—is...
$$y(t) = C_1e^{\frac{-1 + 7i}{5} t} + C_2e^{\frac{-1 - 7i}{5} t} \\ \\ = C_1e^{\frac{-1}{5}t}e^{\frac{7i}{5}t} + C_2e^{\frac{-1}{5}t}e^{\frac{-7i}{5}t} \\ = e^{\frac{-1}{5}t} (C_1e^{\frac{7i}{5}t} + C_2e^{\frac{-7i}{5}t}) \\ = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t}) \\ = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t})$$
Since cos[x] is even and sin[x] is odd...
$$= e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{7}{5}t} - C_2isin{\frac{7}{5}t}) \\ = e^{\frac{-1}{5}t} ( (C_1 + C_2)cos{\frac{7}{5}t} + (C_1 - C_2)isin{\frac{7}{5}t}) \\ \text{Let:} \\ A = (C_1 + C_2), B = i(C_1 - C_2) \\ \text{So...} \\ y(t) = e^{\frac{-1}{5}t} (Acos{\frac{7}{5}t} + Bsin{\frac{7}{5}t})$$

Now we use the initial conditions to solve for A and B.

$$y(0) = e^{\frac{-1}{5}(0)} (A\cos{\frac{7}{5}(0)} + B\sin{\frac{7}{5}(0)}) = 1 \\ = A\cos 0 + B\sin 0 = 1 \\ A = 1 \\ y'(t) = \frac{-1}{5}e^{\frac{-1}{5}t}(A\cos{\frac{7}{5}t} + B\sin{\frac{7}{5}t}) + e^{\frac{-1}{5}t} (\frac{-7}{5}A\sin{\frac{7}{5}t} + \frac{7}{5}B\cos{\frac{7}{5}t})$$

We end up w/: B = 1/7

So...
$$y(t) = e^{\frac{-1}{5}t} \left(\cos \frac{7}{5}t + \frac{1}{7}\sin \frac{7}{5}t \right)$$

(I'm not sure why the LaTeX stuff is showing up in boxes like that; I've tried a few things but it's always coming up).

Last edited by a moderator: Mar 31, 2013
2. Mar 31, 2013

### Zondrina

$y(t) = c_1e^{\frac{-1 + 7i}{5} t} + c_2e^{\frac{-1 - 7i}{5} t} \\ = c_1e^{-\frac{t}{5}}(cos(\frac{7t}{5}) + isin(\frac{7t}{5})) + c_2e^{-\frac{t}{5}}(cos(\frac{7t}{5}) - isin(\frac{7t}{5}))$

Find y'.

Now use those initial conditions.

3. Mar 31, 2013

### vela

Staff Emeritus
Wolfram Alpha's answer agrees with yours. It could be you're entering the answer incorrectly if it's graded by a computer. I expect, however, it's because the coefficient of y'' in your differential equation is wrong. It's suppose to be the mass, not the weight of the object.