# From equation $\frac{d^2\vec{r}}{dt^2} + 4\vec{r} = \vec{0}$, solve $\vec r$

The motion of a particle is described by the position vector $$\vec{r}$$ at time $$t$$, following the equation $$\frac{d^2\vec{r}}{dt^2} + 4\vec{r} = \vec{0}$$. Determine the expression for $$\vec{r}$$ at any time $$t$$ under the given conditions $$\vec{r} = \vec{a}$$ and $$\frac{dr}{dt} = 0$$ at $$t = 0$$.

Sol: Integrating the differential equation $$(\frac{dr}{dt})^2+4\vec r^2=c$$. Using IC, $$c=4\vec r^2$$.

Then $$\frac{dr}{dt}=2\sqrt{\vec a^2-\vec r^2}$$.

Integrating we get $$\sin^{-1}|\frac{\vec r}{\vec a}|=2t+c'$$ and using IC ($$\vec r = \vec a$$ when $$t = 0$$),

we get $$|\vec r|=|\vec a| \cos 2t.$$

Then, $$\vec r$$ and $$\vec a$$ are parallel vectors. If this can be written, I can easily solve the next part. But I am aware that if two vectors are not parallel, they could have same magnitudes but different directions.

Is the conclusion, Then, $$\vec r$$ and $$\vec a$$ are parallel vectors correct? If yes, why? How to solve this problem, if the above approach is not correct? Please help.

Edit: If $$\vec r$$ and $$\vec a$$ are parallel vectors, then the unit vectors along them. Then $$\vec r=\vec a\frac{|\vec r|}{|\vec a|}=\vec a cos(2t)$$. Is it possible to provide a better argument to show $$\vec r$$ and $$\vec a$$ are parallel vectors?

By solving the ODE componentwisely, we have $$r_i=c_1\cos(2t)+c_2\sin(2t)$$ for $$1\le i\le3$$, and to rewrite it in vector form, its general solution is simply $$\vec r = \cos(2t) \vec c_1 + \sin(2t) \vec c_2$$
By $$\vec r(0)=\vec a$$, we get $$\vec c_1 = \vec a$$; then $$\vec r'(0)=0$$, we get $$\vec c_2=\vec 0$$. Therefore, $$\vec r = \cos(2t) \vec a$$
• Thanks. Please see the edit. If I want to explore the method I have provided, who to make that errorless? Is there any argument to show $\vec r$ and $\vec a$ are parallel vectors? Dec 11, 2023 at 8:28
• I don't know much rigor you need, but this is pretty obvious from a mechanics point of view. Both of the initial velocity and replacement are scalar multiples of $\vec a$, as well as the acceleration (which is $-4\vec r$), so there is no way for the particle to move out of the 1-dimensional space it started with. It's really just a simple harmonic system, such as a particle attached to a spring. Dec 11, 2023 at 9:10