Question

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Answer 1

(a)

Take equation (9), the total energy E is

     $E=\frac{1}{2}[kx^2+m(x')^2]$

Differentiating E with respect to t,

    $E'(t)=\frac{1}{2}[k(2xx')+2m(x')(x'')]$

                $=\frac{1}{2}(2x')[kx+mx'']=x'[kx+mx'']$

Using equation (1),

      $mx''+cx'+kx=f(t)\Rightarrow mx''+kx=f(t)-cx'$

Then

       $E'(t)=x'[f(t)-cx']$ which is our required differential equation.

(b)

Assume that f(t)=0 implies,

       $E'(t)=x'[-cx']$

                   $=-c(x')^2$

for constant energy, $\frac{d}{dt}E(t))=0\Rightarrow c(x')^2=0$

                                                             $\Rightarrow c=0.$

Then the condition on c=0 guarantees that the energy in the system is constant.

                $\frac{d}{dt}E(t)=0$

(c)

Referring part(a), Assume $c\neq 0,$

$E'(t)=x'[f(t)-cx']=0$

    $x'=0 or f(t)-cx'=0$

                      $f(t)=cx'$

(d)

Plugging in f(t)=0, m=1, c=2 , k=2, x(0)=0, x'(0)=1

then equation (1) becomes,

$x''+2x'+2x=0, x(0)=0, x'(0)=1$

the auxiliary equation is m^2+2m+2=0

Then the solution is $m=\frac{-2\pm \sqrt{4-4(1)(2)}}{2}= \frac{-2\pm 2i}{2}=-1\pm i$

Then the solution of this equation is

       $x(t)=e^{-t}\left [{Acos(t)+Bsin(t)} \right ]$

Now applying the initial condition,

$x(0)=0 \Rightarrow x(0)=e^{0}\left [{Acos(0)+Bsin(0)} \right ] \Rightarrow A=0$

$x'(t)=-e^{-t}[Acos(t)+Bsin(t)]+e^{-t}[-Asin(t)+Bcos(t)]$

$x'(0)=1\Rightarrow x'(0)=-e^{-0}[Acos(0)+Bsin(0)]+e^{-0}[-Asin(0)+Bcos(0 )]$

This implies 1=B

Then $x(t)=e^{-t}sin(t)$