Solving Harmonic Oscillator Equation w/ Initial Conditions

[quas]

Homework Statement

a mass is placed on a loose spring and connected to the ceiling. the spring is connected to the floor in t=0 the wire is cut
a. find the equation of the motion
b. solve the equation under the initial conditions due to the question

Homework Equations

$\sum F=ma$
$x(t)=Asin(\omega t + \phi )$

The Attempt at a Solution

a. due to the 2 law of Newton: $\sum F=ma_x$
$mg-kx=ma$
$mg-kx=m\ddot{x} \\ \ddot{x}=g-\frac{k}{m}x$

b. first I'll find the point equilibrium
$c-kx=ma \\ c-kx=m\cdot 0 \\ x_0=\frac{c}{k}$
then I'll define $y=x-x_0$

How do I go from here?
thanks

 

[TSny]

$\ddot{x}=g-\frac{k}{m}x$

OK

$x_0=\frac{c}{k}$

What does c stand for?

then I'll define $y=x-x_0$

How do I go from here?

Rewrite the differential equation in terms of y instead of x.

 

[Cutter Ketch]

As TSny hints, you tripped finding x0. Think again what condition will be true at equilibrium and define x0 again. There won't be an unknown c.

 

[quas]

OK

What does c stand for?Rewrite the differential equation in terms of y instead of x.

sorry I meant $x_0 = \frac{mg}{k}$
ok and then $\ddot{y}=g-\frac{k}{m}y$ ?

 

[BvU]

$\ddot{y}=g-\frac{k}{m}y$ can't be right. Compare with $\ddot{x}=g-\frac{k}{m}x$

 

[quas]

$\ddot{y}=g-\frac{k}{m}y$ can't be right. Compare with $\ddot{x}=g-\frac{k}{m}x$

I might have a barrier but I do not understand how to build a differential equation for y,,,,
I understand that $\dot{x}=\frac{dx}{dt}=\frac{dy}{dt}=\dot{y} \\ a=\ddot{x}=\frac{d^2x}{dt^2}=\frac{d^2y}{dt^2}=\ddot{y}$

 

[Cutter Ketch]

$\ddot{y}=g-\frac{k}{m}y$ can't be right. Compare with $\ddot{x}=g-\frac{k}{m}x$

? I must be thick this morning, but if positive y is down that looks ok to me.

 

[Cutter Ketch]

? I must be thick this morning, but if positive y is down that looks ok to me.

Oh, good grief. It took me a minute!

 

[Cutter Ketch]

I might have a barrier but I do not understand how to build a differential equation for y,,,,
I understand that $\dot{x}=\frac{dx}{dt}=\frac{dy}{dt}=\dot{y} \\ a=\ddot{x}=\frac{d^2x}{dt^2}=\frac{d^2y}{dt^2}=\ddot{y}$

Much later after it stops oscillating and y is y0 what is the acceleration? g?

 

[BvU]

I might have a barrier but I do not understand how to build a differential equation for y,,,,
I understand that $\dot{x}=\frac{dx}{dt}=\frac{dy}{dt}=\dot{y} \\ a=\ddot{x}=\frac{d^2x}{dt^2}=\frac{d^2y}{dt^2}=\ddot{y}$

Simple: the second derivatives on the left are equal alright. But you need to substitute your expression for y in terms of x. Not just y = x, but: ...