A spring-mass system has natural frequency 7sqrt(10) rad/s. The natural length of the spring is 0.7 m. Find the length of the spring when the mass is in equilibrium.

Question

idealist10 · Answer

$$y''+\frac{ k }{ m }y=0$$

idealist10 · Answer

@freckles @jim_thompson5910 @Kainui @mathstudent55 @SolomonZelman @Zarkon

solomonzelman · Answer

Undamped free case:)

solomonzelman · Answer

Let  $\color{#000000}{ \displaystyle \omega:=\sqrt{\frac{k}{m}}}$

$\color{#000000}{ \displaystyle y''+\frac{k}{m}y=0\quad \Rightarrow \quad y''+\omega^2y=0}$

$\color{#000000}{ \displaystyle r^2+\omega^2=0}$
$\color{#000000}{ \displaystyle r=\pm\omega{\tiny~} i}$

and then the "prove" for a simple harmonica motion formula
(by letting $c_1=A\sin\phi$  and  $c_2=A\cos\phi$, and re-writing)

but, you don't want to discuver formulas, just solve it, right?

solomonzelman · Answer

You can write the auxiliary equation
$\color{#000000}{ \displaystyle r^2+\frac{k}{m}=0}$      ===>      $\color{#000000}{ \displaystyle r=\pm i\sqrt{\frac{k}{m}}}$

what is exactly the problem tho?

idealist10 · Answer

So suppose that |dw:1462406736850:dw|