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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?

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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, $$\vec r = \cos(2t) \vec a$$

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  • $\begingroup$ 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? $\endgroup$ Dec 11, 2023 at 8:28
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    $\begingroup$ 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. $\endgroup$ Dec 11, 2023 at 9:10

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