2
$\begingroup$

Suppose a system containing a mass $m$ on frictionless surface, attached by a spring to a wall. The spring's constant is complex, given by $K = K_1 + K_2i$, with $K_1 \gg K_2$. Write the equation of motion, and show it has the dynamic of a damped oscillator.

So, Newton's second law: $$ m\ddot{x} = -(K_1+K_2i)x \Leftrightarrow \ddot{x} = -\frac{K_1+K_2i}{m}x $$ Solve for $x = e^{\lambda t}$, we get $$ \lambda^2 = -\frac{K_1 + K_2i}{m} \Leftrightarrow \lambda = \pm\frac{1}{\sqrt{m}}\sqrt{-(K_1+K_2i)} $$ $$ \lambda = \pm\sqrt{\frac{K_1}{m}}i\sqrt{1+\frac{K_2}{K_1}i}\approx \pm\sqrt{\frac{K_1}{m}}i\left(1+\frac{K_2}{2K_1}i\right) = \pm\sqrt{\frac{K_1}{m}}i\mp\sqrt{\frac{K_1}{m}}\frac{K_2}{2K_1} $$ This is where I'm stuck. Since the roots are not complex conjugates, we don't get the solution $$x(t) = e^{-\sqrt{\frac{K_1}{m}}\frac{K_2}{2K_1}t}\left(A\cos\left(\sqrt{\frac{K_1}{m}}t\right) + B\sin\left(\sqrt{\frac{K_1}{m}}t\right)\right)$$

Which is what I was guided to find. The solutions will be complex, which is not physical.

Any help?

$\endgroup$

2 Answers 2

1
$\begingroup$

I'm not a physics guru (yet!), but here's my 50 cents:

Since $\lambda$ represents complex vibration frequency, you're not interested in a negative real part of the root (as negative frequency isn't physical, or at least in the naive interpretation).

So instead, you're taking the positive root, and then if $\lambda$ is a solution to the ODE, so is its conjugate.

$\endgroup$
0
$\begingroup$

The most general solution to the equation is $x = c_1 e^{-\lambda t} + c_2 e^{\lambda t}$, where the constants are complex. The physical choice for lambda in each of these terms is the one that produces exponential decay. So, plugging in the proper values of $\lambda$ gives you $$x = e^{-\sqrt{\frac{K_1}{m}}\frac{K_2}{2 K_1}t}\left(c_1 e^{i\sqrt{\frac{K_1}{m}}t} + c_2 e^{-i\sqrt{\frac{K_1}{m}t}}\right)$$ Then you can redefine the complex constants as $c_1 = (A-iB)/2$ and $c_2 = (A+iB)/2$ and use some identities to get the equation in the form that you put up above. I'd note that the form with complex coefficients and complex exponentials is the most general form, but the form you were trying to find is the general form of real solutions to the equations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.