There is an object moving in circle, with this law:

$$ \alpha = -k^2 \theta $$

With $ \alpha = \frac{ d \omega }{dt} $, and $ \omega= \frac{d \theta}{dt} $, $\theta$ = angle, $k$ positive constant.

I need to compute $ \theta $ in function of time. My attempt:

$$ \frac{d^2 \theta}{dt^2} = -k \theta $$ $$ \frac{d^2 \theta}{\theta} = -k^2 dt^2 $$ $$ \int_{\theta_0}^{\theta} {\frac{d^2 \theta}{\theta}} = - \int{k^2 dt^2} $$ $$ \ln\Big(\frac{\theta}{\theta_0}\Big) d\theta = -k^2 t \cdot dt$$

So I try to get the solution computing:

$$ \int_{\theta_0}^{\theta} \ln\Big(\frac{\theta}{\theta_0}\Big) d\theta = -\int k^2 t \cdot dt $$

But it's wrong, the solution is $\theta = \theta_0 \sin \big( kt + \phi \big) $.

  • $\begingroup$ You face a standard differential equation, which is $y'' +\omega^2 y =0$. If $\omega \in \mathbb{R}$ then one way to write the general solution of this equation is $y(t) = \lambda \sin(\omega t + \phi)$ where $\lambda$ and $\phi$ are two constants that depends on initial conditions. You can check it yourself since you only need to derive $y(t)$ two times. But the way you started seems a bit complicated to me, the equation you're resolving is to simple to do that. $\endgroup$
    – ChocoPouce
    Commented Apr 17, 2014 at 13:23
  • $\begingroup$ But $ \theta_0 \sin \big( kt + \phi \big) $ derived two times is $ - \theta_0 k^2 \sin \big( kt + \phi \big) $, not $-k^2 \theta$. $\endgroup$ Commented Apr 17, 2014 at 13:45
  • $\begingroup$ Since $\theta = \theta_0 \sin(kt+\phi)$ I think you are ok. $\endgroup$
    – ChocoPouce
    Commented Apr 17, 2014 at 13:48

2 Answers 2


The equation is a second-order linear differential equation; in standard form,

$$\frac{\mathrm{d}^2 \theta(t)}{\mathrm{d}t^2} + k^2\theta = 0$$

For the case of constant coefficients, one must simply propose the ansatz $\theta(t)=e^{Rt}$, where $R$ is a constant. Plugging into the differential equation yields,

$$e^{Rt}(R^2+k^2) = 0$$

This is only the case when $R^2+k^2=0$, which has solutions $R=\pm i k$. Using Euler's formula, we can express the two solutions of the differential equation as,

$$\theta_1 = \cos(kt) + i\sin(kt), \, \, \, \, \theta_2 = \cos(kt) - i\sin(kt)$$

However, recall the superposition principle applies to this differential equation, and hence any linear combination of these solutions is also a solution. We construct the following new solutions:

$$\Theta_1 = \frac{1}{2}(\theta_1 + \theta_2) = \cos(kt)$$ $$\Theta_2 = \frac{1}{2i}(\theta_1-\theta_2) = \sin(kt)$$

We can once again combine these solutions, and introduce two arbitrary constants (determined by initial conditions) to propose the general solution,

$$\theta(t) = c_1 \cos(kt) + c_2\sin(kt)$$

The equation is well-known, and that of a classical harmonic oscillator. Any calculus text should offer a treatment of a second-order ODE with constant coefficients. See http://tutorial.math.lamar.edu/ for free online calculus resources. The free course http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/ also offers a thorough introduction to differential equations.


If you don't understand the other method, this is how you would do in the realm of real numbers :

Write $$\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\omega\frac{d\omega}{d\theta}$$

I have used that $\omega=\frac{d\theta}{dt}$ in the last step. Now your equation is first order and you can solve it easily.

$$\omega d \omega=-k^2 \theta d\theta$$

You can take from here I guess.


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