# Harmonic oscillator

## 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
Homework Helper
Gold Member
##\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.

• quas
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 ## ?

Last edited:
BvU
Homework Helper
##
\ddot{y}=g-\frac{k}{m}y## can't be right. Compare with ##
\ddot{x}=g-\frac{k}{m}x##

##
\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} ##

##
\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.

??? 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!!

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
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: ... 