Answer to Question #105990 in Differential Equations for Pappu Kumar Gupta

Question

Interpret the initial value problem
0
0
0
2
2
2
0, (0) , w
q
b q q q
q
=


+ = =
t= dt
d
dt
d
for any physical situation and hence solve the problem

Answer 1

Interpret the initial value problem

$${d^2 \theta \over dt^2}+\beta^2\theta=0, \ \theta(0)=\theta_0,\ {d\theta \over dt}(0)=\omega_0$$

Use the notation

$${d^2 \theta \over dt^2}=\theta'',\ {d \theta \over dt}=\theta'$$

We get

$$\theta''+\beta^2\theta=0$$

The auxiliary equation is $r^2+\beta^2=0$ or $r^2=-\beta^2,$ whose roots are $\pm i\beta.$

The general solution is

$$\theta(t)=c_1 \cos(\beta t)+c_2\sin(\beta t)$$

Then

$$\theta'=-\beta c_1\sin(\beta t)+\beta c_2\cos(\beta t)$$

The initial conditions become

$$\theta(0)=c_1=\theta_0$$

$$\theta'(0)=\beta c_2=\omega_0=>c_2={\omega_0 \over \beta}$$

Hence

$$\theta(t)=\theta_0 \cos(\beta t)+{\omega_0 \over \beta}\sin(\beta t)$$

Spring system

$$\begin{matrix} x & displacement \\ dx/dt & velocity \\ m & mass \\ c & damping\ constant \\ k & spring\ constant \\ F(t) & external\ force \end{matrix}$$

If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals mass times acceleration), we have

$$m{d^2 x \over dt^2}=-kx$$

Or

$${d^2 x \over dt^2}+{k\over m}x=0$$

Let $\beta^2=\dfrac{k}{m}.$ Then the solution of the initial value problem

$${d^2 x \over dt^2}+\beta^2x=0, \ x(0)=x_0,\ {dx\over dt}(0)=v_0$$

is

$$x(t)=x_0 \cos(\beta t)+{v_0 \over \beta}\sin(\beta t), \beta=\sqrt{{k \over m}}$$