Why can all solutions to the simple harmonic motion equation be written in terms of sines and cosines?












3












$begingroup$


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.










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$endgroup$








  • 3




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
















3












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 3




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














3












3








3


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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edited 1 hour ago









knzhou

44.3k11121213




44.3k11121213










asked 2 hours ago









ADITYA PRAKASHADITYA PRAKASH

615




615








  • 3




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














  • 3




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








3




3




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




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










2 Answers
2






active

oldest

votes


















9












$begingroup$

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. Because of this, I (and many others) generally refer to this as the Method of Divine Inspiration: I can just tell you that the solution came to me in a dream, handed over by a flying mass of spaghetti, and $-$ no matter how contrived or elaborate the solution looks $-$ if it passes the two criteria above, the fact that it is the solution is bulletproof, and no further explanation of how it was built is required.



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











$endgroup$









  • 1




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










  • $begingroup$
    @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
    $endgroup$
    – Emilio Pisanty
    2 hours ago










  • $begingroup$
    I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
    $endgroup$
    – alephzero
    1 hour ago








  • 1




    $begingroup$
    @alephzero There are no "better" answers - that's the whole point.
    $endgroup$
    – Emilio Pisanty
    48 mins ago










  • $begingroup$
    @alephzero How is it better...?
    $endgroup$
    – Aaron Stevens
    41 mins ago



















2












$begingroup$


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 soon just be able to notice them. It's the like 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 solving such equations when the solutions are canonical.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    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. Because of this, I (and many others) generally refer to this as the Method of Divine Inspiration: I can just tell you that the solution came to me in a dream, handed over by a flying mass of spaghetti, and $-$ no matter how contrived or elaborate the solution looks $-$ if it passes the two criteria above, the fact that it is the solution is bulletproof, and no further explanation of how it was built is required.



    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











    $endgroup$









    • 1




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










    • $begingroup$
      @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
      $endgroup$
      – Emilio Pisanty
      2 hours ago










    • $begingroup$
      I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
      $endgroup$
      – alephzero
      1 hour ago








    • 1




      $begingroup$
      @alephzero There are no "better" answers - that's the whole point.
      $endgroup$
      – Emilio Pisanty
      48 mins ago










    • $begingroup$
      @alephzero How is it better...?
      $endgroup$
      – Aaron Stevens
      41 mins ago
















    9












    $begingroup$

    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. Because of this, I (and many others) generally refer to this as the Method of Divine Inspiration: I can just tell you that the solution came to me in a dream, handed over by a flying mass of spaghetti, and $-$ no matter how contrived or elaborate the solution looks $-$ if it passes the two criteria above, the fact that it is the solution is bulletproof, and no further explanation of how it was built is required.



    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











    $endgroup$









    • 1




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










    • $begingroup$
      @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
      $endgroup$
      – Emilio Pisanty
      2 hours ago










    • $begingroup$
      I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
      $endgroup$
      – alephzero
      1 hour ago








    • 1




      $begingroup$
      @alephzero There are no "better" answers - that's the whole point.
      $endgroup$
      – Emilio Pisanty
      48 mins ago










    • $begingroup$
      @alephzero How is it better...?
      $endgroup$
      – Aaron Stevens
      41 mins ago














    9












    9








    9





    $begingroup$

    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. Because of this, I (and many others) generally refer to this as the Method of Divine Inspiration: I can just tell you that the solution came to me in a dream, handed over by a flying mass of spaghetti, and $-$ no matter how contrived or elaborate the solution looks $-$ if it passes the two criteria above, the fact that it is the solution is bulletproof, and no further explanation of how it was built is required.



    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











    $endgroup$



    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. Because of this, I (and many others) generally refer to this as the Method of Divine Inspiration: I can just tell you that the solution came to me in a dream, handed over by a flying mass of spaghetti, and $-$ no matter how contrived or elaborate the solution looks $-$ if it passes the two criteria above, the fact that it is the solution is bulletproof, and no further explanation of how it was built is required.



    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














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 hours ago

























    answered 2 hours ago









    Emilio PisantyEmilio Pisanty

    83.5k22204419




    83.5k22204419








    • 1




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










    • $begingroup$
      @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
      $endgroup$
      – Emilio Pisanty
      2 hours ago










    • $begingroup$
      I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
      $endgroup$
      – alephzero
      1 hour ago








    • 1




      $begingroup$
      @alephzero There are no "better" answers - that's the whole point.
      $endgroup$
      – Emilio Pisanty
      48 mins ago










    • $begingroup$
      @alephzero How is it better...?
      $endgroup$
      – Aaron Stevens
      41 mins ago














    • 1




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










    • $begingroup$
      @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
      $endgroup$
      – Emilio Pisanty
      2 hours ago










    • $begingroup$
      I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
      $endgroup$
      – alephzero
      1 hour ago








    • 1




      $begingroup$
      @alephzero There are no "better" answers - that's the whole point.
      $endgroup$
      – Emilio Pisanty
      48 mins ago










    • $begingroup$
      @alephzero How is it better...?
      $endgroup$
      – Aaron Stevens
      41 mins ago








    1




    1




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




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












    $begingroup$
    @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
    $endgroup$
    – Emilio Pisanty
    2 hours ago




    $begingroup$
    @Frotaur That's an interesting name for the Picard-Lindelöf theorem ;-).
    $endgroup$
    – Emilio Pisanty
    2 hours ago












    $begingroup$
    I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
    $endgroup$
    – alephzero
    1 hour ago






    $begingroup$
    I still remember the favourite catch-phrase of the lecturer in my ODE course: "They way to solve this equation is to look at it until you see what the solution is". But actually, the OP's equation is linear with constant coefficients, and the general theory of how to solve them is well known. A better answer would be "because $e^{i omega t} = cos omega t + i sin omega t$."
    $endgroup$
    – alephzero
    1 hour ago






    1




    1




    $begingroup$
    @alephzero There are no "better" answers - that's the whole point.
    $endgroup$
    – Emilio Pisanty
    48 mins ago




    $begingroup$
    @alephzero There are no "better" answers - that's the whole point.
    $endgroup$
    – Emilio Pisanty
    48 mins ago












    $begingroup$
    @alephzero How is it better...?
    $endgroup$
    – Aaron Stevens
    41 mins ago




    $begingroup$
    @alephzero How is it better...?
    $endgroup$
    – Aaron Stevens
    41 mins ago











    2












    $begingroup$


    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 soon just be able to notice them. It's the like 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 solving such equations when the solutions are canonical.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$


      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 soon just be able to notice them. It's the like 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 solving such equations when the solutions are canonical.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$


        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 soon just be able to notice them. It's the like 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 solving such equations when the solutions are canonical.






        share|cite|improve this answer











        $endgroup$




        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 soon just be able to notice them. It's the like 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 solving such equations when the solutions are canonical.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 2 hours ago









        ChairChair

        4,26672137




        4,26672137






























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