Answer to Question #148536 in Differential Equations for Nikhil

Question

The motion of a certain spring-mass system is governed by the differential equation d^2u/dt^2+1/8du/dt+u=0 where u is measured in metres and t in seconds. If u(0)=2 and (du/dt)t=0=0 determine the subsequent motion. Also determine the time at which the mass first passes through the equilibrium position.

Answer 1

Let us solve the equation:

$\frac{d^2u}{dt^2}+\frac{1}{8}\frac{du}{dt}+u=0$

The characteeristic equation $k^2+\frac{1}{8}k+1=0$ is equivalent to $(k+\frac{1}{16})^2=\frac{1}{256}-1=-\frac{255}{256}$

So, $k+\frac{1}{16}=\pm i\frac{\sqrt{255}}{16}$ and thus $k=\frac{-1\pm i\sqrt{255}}{16}$.

Therefore, the solution of the differential equation is

$u=e^{-\frac{t}{16}}(C_1\cos(\frac{\sqrt{255}}{16} t)+C_2\sin(\frac{\sqrt{255}}{16} t))$

Then $\frac{du}{dt}=-\frac{1}{16}e^{-\frac{t}{16}}(C_1\cos(\frac{\sqrt{255}}{16} t)+C_2\sin(\frac{\sqrt{255}}{16} t))+e^{-\frac{t}{16}}(-C_1\frac{\sqrt{255}}{16}\sin(\frac{\sqrt{255}}{16} t)+C_2\frac{\sqrt{255}}{16}\cos(\frac{\sqrt{255}}{16} t))$

If $u(0)=2$ and $\frac{du}{dt}(0)=0$ then we have $2=u(0)=C_1$ and $0=\frac{du}{dt}(0)=-\frac{1}{16}C_1+C_2\frac{\sqrt{255}}{16}$ .

Therefore, $C_1=2$ and $C_2=\frac{2}{\sqrt{255}}$ and the subsequent motion is the following:

$u=e^{-\frac{t}{16}}(2\cos(\frac{\sqrt{255}}{16} t)+\frac{2}{\sqrt{255}}\sin(\frac{\sqrt{255}}{16} t))= 2e^{-\frac{t}{16}}(\cos(\frac{\sqrt{255}}{16} t)+\frac{1}{\sqrt{255}}\sin(\frac{\sqrt{255}}{16} t)).$

Let us determine the time at which the mass first passes through the equilibrium position. For this we should solve the equation:

$u(t)=0$

$2e^{-\frac{t}{16}}(\cos(\frac{\sqrt{255}}{16} t)+\frac{1}{\sqrt{255}}\sin(\frac{\sqrt{255}}{16} t))=0$

$\cos(\frac{\sqrt{255}}{16} t)+\frac{1}{\sqrt{255}}\sin(\frac{\sqrt{255}}{16} t)=0$

$\tan(\frac{\sqrt{255}}{16} t)=-\sqrt{255}$

$\frac{\sqrt{255}}{16} t\approx -1.51+3.14 n, \ n\in\mathbb Z.$

Since the mass first passes through the equilibrium position and $t\ge 0$, we conclude that $n=1$. Therefore, $\frac{\sqrt{255}}{16} t\approx -1.51+3.14=1.63$ and $t\approx 1.63\frac{16}{\sqrt{255}}\approx 1.633$ (seconds).