Answer to Question #125357 in Differential Equations for jse

Question

1. A mass of 250 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 20 cms, and if there is no damping, determine the position u of the mass at any time t.

Enclose arguments of functions in parentheses. For example, sin(2x).

a. Assume g = 9.8 m/s^2. Enter ans exact answer.
u(t) = ______ m

b. When does the mass first return to its equilibrium position?
Enter an exact answer.

t = ______s

Answer 1

$m=0.25$ kg

$L=0.01568$ m

$u(0)=0, \ u^{\prime}(0)=0.2$ m/s, where $u(t)$ is position of the mass at any time $t.$

$k=\frac{mg}{L}=\frac{0.25\times 9.8}{0.01568}=156.25$ N/m

Now we can write the equation of the system: $mu^{\prime \prime }+ku =0$

$0.25u^{\prime \prime }+156.25u =0$

$u^{\prime \prime }+625u =0$

The characteristic equation: $r^2+625=0, \ \ r=\pm25i$ .

The general solution is $\ u(t)=C_1\sin(25t)+C_2\cos(25t)$

$u(0)=0=C_1\times 0+C_2\times 1=C_2$ , $C_2=0$

$u^{\prime }(0)=0.2=25C_1$ , $C_1=0.008$

So, $u(t)=0.008\sin (25t)$ m.

Equilibrium position: $u(t)=0, \ \ 0.008\sin(25t)=0, \ \ \sin(25t)=0, \ \ 25t=\pi n, \ n\in \mathbb{Z}$ .

The mass first return to its equilibrium position when $n=1$ or $t=\pi/25.$

Answer:

a. $u(t)=0.008\sin (25t)$ m.

b. $t=\pi/25$ s.