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Why can SHM equations be described in sin(s) and cosine(s)?

Which trigonometric ratio should be used to describe simple harmonic motion as a function of time?Pendulum's motion is simple harmonic motionPhase angle in simple harmonic motionSimple harmonic motion about an unstable equilibrium position?Simple harmonic motion confusionCan a periodic motion whose displacement is given by $ x=sin^2(omega t)$, be considered as a SHM?Intuitively why time period of shm do not depend on displacement from mean position, ie. amplitudeWhile determining phase, is it okay to choose displacement as zero?What is the difference between Non-harmonic oscillation, Anharmonic oscillation and Complex harmonic oscillation?Which equation to use for SHM?

2

$begingroup$

Defining property of SHM (simple harmonic motion)-force experienced at any value of displacement from mean position is directly proportional to it and is directed towards mean position i.e $F=-k(x)$.

From this,
$$mleft(frac{d^2x}{dt^2}right) +kx=0.$$

Then I read from this site

Let us interpret this equation. The second derivative of a function of x plus the function itself (times a constant) is equal to zero. Thus the second derivative of our function must have the same form as the function itself. What readily comes to mind is the sine and cosine function.

How can we assume so plainly that it should be sin or cosine only , they do satisfy the equation but why are they brought in the picture so directly, what I want to ask is why can SHM displacement, velocity etc. be expressed in sin and cosine, I know the "SHM is projection of uniform circular motion" proof, but an algebraic proof would be appreciated.

harmonic-oscillator

share|cite|improve this question

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac{mathrm{d}^2x}{mathrm{d}t}$ missing a power of 2?
  $endgroup$
  – Chair
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How many solutions $f$ are there to $f''=-lambda f$?
  $endgroup$
  – Kyle Kanos
  26 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Defining property of SHM (simple harmonic motion)-force experienced at any value of displacement from mean position is directly proportional to it and is directed towards mean position i.e $F=-k(x)$.

From this,
$$mleft(frac{d^2x}{dt^2}right) +kx=0.$$

Then I read from this site

Let us interpret this equation. The second derivative of a function of x plus the function itself (times a constant) is equal to zero. Thus the second derivative of our function must have the same form as the function itself. What readily comes to mind is the sine and cosine function.

How can we assume so plainly that it should be sin or cosine only , they do satisfy the equation but why are they brought in the picture so directly, what I want to ask is why can SHM displacement, velocity etc. be expressed in sin and cosine, I know the "SHM is projection of uniform circular motion" proof, but an algebraic proof would be appreciated.

harmonic-oscillator

share|cite|improve this question

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac{mathrm{d}^2x}{mathrm{d}t}$ missing a power of 2?
  $endgroup$
  – Chair
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How many solutions $f$ are there to $f''=-lambda f$?
  $endgroup$
  – Kyle Kanos
  26 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Defining property of SHM (simple harmonic motion)-force experienced at any value of displacement from mean position is directly proportional to it and is directed towards mean position i.e $F=-k(x)$.

From this,
$$mleft(frac{d^2x}{dt^2}right) +kx=0.$$

Then I read from this site

Let us interpret this equation. The second derivative of a function of x plus the function itself (times a constant) is equal to zero. Thus the second derivative of our function must have the same form as the function itself. What readily comes to mind is the sine and cosine function.

How can we assume so plainly that it should be sin or cosine only , they do satisfy the equation but why are they brought in the picture so directly, what I want to ask is why can SHM displacement, velocity etc. be expressed in sin and cosine, I know the "SHM is projection of uniform circular motion" proof, but an algebraic proof would be appreciated.

harmonic-oscillator

share|cite|improve this question

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

$endgroup$

share|cite|improve this question

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

share|cite|improve this question

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

edited 25 mins ago

Emilio Pisanty

83.4k22204419

edited 25 mins ago

Emilio Pisanty

83.4k22204419

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

asked 37 mins ago

ADITYA PRAKASHADITYA PRAKASH

565

2 Answers
2

active

oldest

votes

5

$begingroup$

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions.

The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can construct, through whatever means you can come up with, two linearly-independent solutions, then you're guaranteed that any solution of the equation will be a linear combination of your two solutions.

Thus, it doesn't matter at all how it is that you come to the proposal of $sin(omega t)$ and $cos(omega t)$ as prospective solutions: all you need to do is

1. verify that they are solutions, i.e. just plug them into the derivatives and see if the result is identically zero; and


2. check that they're linearly independent.

Once you do that, the details of how you built your solutions become completely irrelevant.

If this framework is unclear or unfamiliar, then you should sit down with an introductory textbook on differential equations. There's a substantial bit of background that makes this sort of thing clearer, and which simply doesn't fit within this site's format.

share|cite|improve this answer

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Might be worth mentioning that this is the "Cauchy-Lipshitz theorem", if OP wants to check it out.
  $endgroup$
  – Frotaur
  12 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

How can we assume so plainly that it should be sin or cosine only

It's literally just a guess. Those're obvious solutions which can be verified easily, and when they're such straightforward functions, you'll just be able to notice them. It's the same way one guesses that when you have an equation like $f'(x)=Ktimes f(x)$, you just see that the solutions are exponentials. After that, you know that for a differential equation like $f^{(n)}(x)=Kf(x)$ you can have upto $n$ solutions, so you aren't missing anything when you consider the sine and cosine.

It's a nice idea to not waste time/effort/space formally integrating such equations when the solutions are canonical. Once you're somewhat familiar with the differential equations, you'll just notice such stuff and not bother to do that integration each time.

share|cite

answered 9 mins ago

ChairChair

4,24672137

$endgroup$

add a comment | 

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5

$begingroup$

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions.

The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can construct, through whatever means you can come up with, two linearly-independent solutions, then you're guaranteed that any solution of the equation will be a linear combination of your two solutions.

Thus, it doesn't matter at all how it is that you come to the proposal of $sin(omega t)$ and $cos(omega t)$ as prospective solutions: all you need to do is

1. verify that they are solutions, i.e. just plug them into the derivatives and see if the result is identically zero; and


2. check that they're linearly independent.

Once you do that, the details of how you built your solutions become completely irrelevant.

If this framework is unclear or unfamiliar, then you should sit down with an introductory textbook on differential equations. There's a substantial bit of background that makes this sort of thing clearer, and which simply doesn't fit within this site's format.

share|cite|improve this answer

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Might be worth mentioning that this is the "Cauchy-Lipshitz theorem", if OP wants to check it out.
  $endgroup$
  – Frotaur
  12 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions.

The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can construct, through whatever means you can come up with, two linearly-independent solutions, then you're guaranteed that any solution of the equation will be a linear combination of your two solutions.

Thus, it doesn't matter at all how it is that you come to the proposal of $sin(omega t)$ and $cos(omega t)$ as prospective solutions: all you need to do is

1. verify that they are solutions, i.e. just plug them into the derivatives and see if the result is identically zero; and


2. check that they're linearly independent.

Once you do that, the details of how you built your solutions become completely irrelevant.

If this framework is unclear or unfamiliar, then you should sit down with an introductory textbook on differential equations. There's a substantial bit of background that makes this sort of thing clearer, and which simply doesn't fit within this site's format.

share|cite|improve this answer

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Might be worth mentioning that this is the "Cauchy-Lipshitz theorem", if OP wants to check it out.
  $endgroup$
  – Frotaur
  12 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions.

The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can construct, through whatever means you can come up with, two linearly-independent solutions, then you're guaranteed that any solution of the equation will be a linear combination of your two solutions.

Thus, it doesn't matter at all how it is that you come to the proposal of $sin(omega t)$ and $cos(omega t)$ as prospective solutions: all you need to do is

1. verify that they are solutions, i.e. just plug them into the derivatives and see if the result is identically zero; and


2. check that they're linearly independent.

Once you do that, the details of how you built your solutions become completely irrelevant.

If this framework is unclear or unfamiliar, then you should sit down with an introductory textbook on differential equations. There's a substantial bit of background that makes this sort of thing clearer, and which simply doesn't fit within this site's format.

share|cite|improve this answer

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

$endgroup$

share|cite|improve this answer

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

answered 19 mins ago

Emilio PisantyEmilio Pisanty

83.4k22204419

0

$begingroup$

How can we assume so plainly that it should be sin or cosine only

It's literally just a guess. Those're obvious solutions which can be verified easily, and when they're such straightforward functions, you'll just be able to notice them. It's the same way one guesses that when you have an equation like $f'(x)=Ktimes f(x)$, you just see that the solutions are exponentials. After that, you know that for a differential equation like $f^{(n)}(x)=Kf(x)$ you can have upto $n$ solutions, so you aren't missing anything when you consider the sine and cosine.

It's a nice idea to not waste time/effort/space formally integrating such equations when the solutions are canonical. Once you're somewhat familiar with the differential equations, you'll just notice such stuff and not bother to do that integration each time.

share|cite

answered 9 mins ago

ChairChair

4,24672137

$endgroup$

add a comment | 

0

$begingroup$

How can we assume so plainly that it should be sin or cosine only

It's literally just a guess. Those're obvious solutions which can be verified easily, and when they're such straightforward functions, you'll just be able to notice them. It's the same way one guesses that when you have an equation like $f'(x)=Ktimes f(x)$, you just see that the solutions are exponentials. After that, you know that for a differential equation like $f^{(n)}(x)=Kf(x)$ you can have upto $n$ solutions, so you aren't missing anything when you consider the sine and cosine.

It's a nice idea to not waste time/effort/space formally integrating such equations when the solutions are canonical. Once you're somewhat familiar with the differential equations, you'll just notice such stuff and not bother to do that integration each time.

share|cite

answered 9 mins ago

ChairChair

4,24672137

$endgroup$

add a comment | 

$begingroup$

How can we assume so plainly that it should be sin or cosine only

It's literally just a guess. Those're obvious solutions which can be verified easily, and when they're such straightforward functions, you'll just be able to notice them. It's the same way one guesses that when you have an equation like $f'(x)=Ktimes f(x)$, you just see that the solutions are exponentials. After that, you know that for a differential equation like $f^{(n)}(x)=Kf(x)$ you can have upto $n$ solutions, so you aren't missing anything when you consider the sine and cosine.

It's a nice idea to not waste time/effort/space formally integrating such equations when the solutions are canonical. Once you're somewhat familiar with the differential equations, you'll just notice such stuff and not bother to do that integration each time.

share|cite

answered 9 mins ago

ChairChair

4,24672137

$endgroup$

share|cite

answered 9 mins ago

ChairChair

4,24672137

answered 9 mins ago

ChairChair

4,24672137

answered 9 mins ago

ChairChair

4,24672137