Studying Simple Harmonic Motion and looking at the derivation.

$ma = \Sigma F$

$m \frac{d^2x}{dt^2} = -kx$

$\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$

The textbook then says that "we want to determine what function of time, $x(t)$, satisfies the equation".

This makes sense from the SHM perspective since we want to define the displacement of a particle in terms of time. However just by looking at the differential equation, how were they able to tell that they had to solve for $x(t)$? Is it simply because there exists the second time derivative of $x$ ?


Let's say I didn't care about $x$ and wanted to solve for the velocity of the particle instead. Would it make sense to do the following:

$\frac{dv}{dt} + \frac{k}{m}x = 0$

and try finding a solution for $v$? (assuming it exists)


You are getting it the wrong way. We never see the equation and then decide to solve for $x$. We need to find $x$ as a function of time and that is why we formulate the equation (which may be solved for determining $x$). Each and every variable in that equation has a physical significance.

$x$ is the only dependent variable here. $x$ in Simple Harmonic Motion represents the position of the particle at any instant. $t$ stands for the time, which as you know is an independent variable. Meanwhile, $m$ (mass) and $k$ are constants. So we need to solve for $x$ which helps us determine the position of the particle at any time. And also using the equation for $x$ we can determine the velocity and acceleration of the particle at any instant by using differentiation.

Read more about Simple Harmonic Motion and its related equations here.

  • $\begingroup$ Thank you for putting it in perspective. It makes sense that the differential was created for the sake of solving for the dependent variable. Please see my followup question in the OP. $\endgroup$ – Carpetfizz Dec 1 '16 at 9:48
  • $\begingroup$ @Carpetfizz For finding velocity, solve for $x$ and then find derivative of $x$ w.r.t.time. Velocity is simple rate of change of $x$ after all. $\endgroup$ – user220382 Dec 1 '16 at 9:49
  • $\begingroup$ @XYZ yes I know that, but for the sake of argument of "formulate the differential equation with determining a value in mind" would my followup equation be a valid setup? I know a general solution for that may not exist. $\endgroup$ – Carpetfizz Dec 1 '16 at 9:51
  • $\begingroup$ @Carpetfizz The follow up equation you wrote is correct, but you cannot solve it for $v$ directly. There are two dependent variables $v$ and $x$. You need to write $v$ as $\frac{dx}{dt}$ in order to solve it. You can solve for only one dependent variable from a single differential equation like that. $\endgroup$ – user220382 Dec 1 '16 at 9:55
  • $\begingroup$ @XYZ makes sense, thank you! $\endgroup$ – Carpetfizz Dec 1 '16 at 9:57

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