Question

Given that $\stackrel{..}{x}+3x=0$, and $x\left(0\right)=1$, $\stackrel{.}{x}\left(0\right)=0,$ whate is $x\left(1\right)$?

• A. $-0.99$
• B. $-0.16$
• C. $0.16$
• D. $0.99$

Answer 1

The correct option is B $-0.16$

$\stackrel{..}{x}+3x=0$ ... (i)
Initial conditions are
$x\left(0\right)=1$
$\stackrel{.}{x}\left(0\right)=0$
Auxiliary equation of (i)
$m^2+3=0$
$m=\pm i\sqrt{3}$
General solution of (i) is
$x\left(t\right)=Acos\sqrt{3}t+Bsin\sqrt{3}t$
Now, $x\left(0\right)=1$
$\Rightarrow 1=A+B.0\Rightarrow A=1$
and $\stackrel{.}{x}\left(t\right)=-A\sqrt{3}sin\sqrt{3}t+B\sqrt{3}cos\sqrt{3}t$
$\stackrel{.}{x}\left(0\right)=0$
$\Rightarrow 0=0+B\sqrt{3}\Rightarrow B=0$
$\therefore x\left(t\right)=cos\sqrt{3}t$
$\therefore x\left(1\right)=cos\sqrt{3}$
$x\left(1\right)=-0.1606$