Question

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 4 inches above the equilibrium position. Find the equation of motion. (Use g

Answer 1

Answer:

The equation of motion is $x(t)=-$$\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    $m\frac{d^2x}{dt} +kx=0$

    $\frac{3}{4} \frac{d^2x}{dt} +72x=0$

  $\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

                                 $m=\sqrt{-96}$

                               =$\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

                                 $x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$  $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

                    $=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

                                                    $c_2$ $4\sqrt{6} =0$

                                     $c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

                    $x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

   $c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is  $-\frac{1}{3} cos4\sqrt{6t}$