Equation of Simple Harmonic Oscillator

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So I know that the differential equation of a simple harmonic oscillator is

$dfracd^2xdt^2=-omega^2x$

and it's solution is $y = Asin(omega t+phi)$.

Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.

In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?

Edit:

Also when I start with the solution itself

$y=Asin(omega t + phi)$

$y'=Aomegacos(omega t + phi)$

$dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$

$omega^2cdot2yy'+2y'y''=0$

$omega^2=-dfracy''y$

$0=dfracyy'''-y''y'y^2$

$yy'''=y''y'$

calculus differential-equations

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edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

add a comment | 

up vote
2
down vote

favorite

So I know that the differential equation of a simple harmonic oscillator is

$dfracd^2xdt^2=-omega^2x$

and it's solution is $y = Asin(omega t+phi)$.

Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.

In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?

Edit:

Also when I start with the solution itself

$y=Asin(omega t + phi)$

$y'=Aomegacos(omega t + phi)$

$dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$

$omega^2cdot2yy'+2y'y''=0$

$omega^2=-dfracy''y$

$0=dfracyy'''-y''y'y^2$

$yy'''=y''y'$

calculus differential-equations

share|cite|improve this question

edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

So I know that the differential equation of a simple harmonic oscillator is

$dfracd^2xdt^2=-omega^2x$

and it's solution is $y = Asin(omega t+phi)$.

Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.

In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?

Edit:

Also when I start with the solution itself

$y=Asin(omega t + phi)$

$y'=Aomegacos(omega t + phi)$

$dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$

$omega^2cdot2yy'+2y'y''=0$

$omega^2=-dfracy''y$

$0=dfracyy'''-y''y'y^2$

$yy'''=y''y'$

calculus differential-equations

share|cite|improve this question

edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

So I know that the differential equation of a simple harmonic oscillator is

$dfracd^2xdt^2=-omega^2x$

and it's solution is $y = Asin(omega t+phi)$.

Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.

In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?

Edit:

Also when I start with the solution itself

$y=Asin(omega t + phi)$

$y'=Aomegacos(omega t + phi)$

$dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$

$omega^2cdot2yy'+2y'y''=0$

$omega^2=-dfracy''y$

$0=dfracyy'''-y''y'y^2$

$yy'''=y''y'$

calculus differential-equations

calculus differential-equations

share|cite|improve this question

edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

share|cite|improve this question

edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

share|cite|improve this question

share|cite|improve this question

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

asked 2 hours ago

Harshit Joshi

319112

asked 2 hours ago

Harshit Joshi

319112

asked 2 hours ago

Harshit Joshi

319112

319112

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote



accepted

Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PeterSzilas Very nice Peter! Thanks a lot
  – gimusi
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Your answer is nice!The link "links" omega and the spring constant.
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
  – Harshit Joshi
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
  – gimusi
  2 hours ago
  

  
  
  
  

 | 
show 7 more comments

up vote
3
down vote

The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PeterSzilas Very nice Peter! Thanks a lot
  – gimusi
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Your answer is nice!The link "links" omega and the spring constant.
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
  – Harshit Joshi
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
  – gimusi
  2 hours ago
  

  
  
  
  

 | 
show 7 more comments

up vote
4
down vote



accepted

Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PeterSzilas Very nice Peter! Thanks a lot
  – gimusi
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Your answer is nice!The link "links" omega and the spring constant.
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
  – Harshit Joshi
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
  – gimusi
  2 hours ago
  

  
  
  
  

 | 
show 7 more comments

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

gimusi

77.6k73889

answered 2 hours ago

gimusi

77.6k73889

answered 2 hours ago

gimusi

77.6k73889

77.6k73889

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PeterSzilas Very nice Peter! Thanks a lot
  – gimusi
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Your answer is nice!The link "links" omega and the spring constant.
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
  – Harshit Joshi
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
  – gimusi
  2 hours ago
  

  
  
  
  

 | 
show 7 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PeterSzilas Very nice Peter! Thanks a lot
  – gimusi
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Your answer is nice!The link "links" omega and the spring constant.
  – Peter Szilas
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
  – Harshit Joshi
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
  – gimusi
  2 hours ago
  

  
  
  
  

1

1

Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
– Peter Szilas
2 hours ago

Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
– Peter Szilas
2 hours ago

@PeterSzilas Very nice Peter! Thanks a lot
– gimusi
2 hours ago

@PeterSzilas Very nice Peter! Thanks a lot
– gimusi
2 hours ago

Gimusi.Your answer is nice!The link "links" omega and the spring constant.
– Peter Szilas
2 hours ago

Gimusi.Your answer is nice!The link "links" omega and the spring constant.
– Peter Szilas
2 hours ago

But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
– Harshit Joshi
2 hours ago

But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
– Harshit Joshi
2 hours ago

@HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
– gimusi
2 hours ago

@HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
– gimusi
2 hours ago

 | 
show 7 more comments

up vote
3
down vote

The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

add a comment | 

up vote
3
down vote

The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

add a comment | 

up vote
3
down vote

up vote
3
down vote

The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Yves Daoust

118k667214

answered 1 hour ago

Yves Daoust

118k667214

answered 1 hour ago

Yves Daoust

118k667214

118k667214

add a comment | 

add a comment | 

 

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