Answer to Question #95276 in Differential Equations for Rajni

Question

A mass m , free to move along a line is attracted towards a given point on the line with a force proportional to its distance from the given point.If the mass starts from rest at a distance x_o from the given point, show that the mass moves in a simple harmonic motion?

Answer 1

Introduce a Cartesian coordinate system, where the center of attraction is origin and $\vec{r}(0)=x_0\vec{e}_1$

We have $m\vec{r}''=-\alpha\vec{r}$ for some $\alpha>0$ . Let $\frac{\alpha}{m}=\omega^2$ , then $\vec{r}''+\omega^2\vec{r}=0$.

If $\vec{r}=(x_1,\ldots,x_n)$, then $x_i''+\omega^2 x_i=0$ for all $i=1,\ldots,n$ .

So for all $i=1,\ldots,n$ we have the general solution $x_i=B_i\cos\omega t+C_i\sin\omega t=$

$\frac{1}{\sqrt{B_i^2+C_i^2}}\bigl(\frac{B_i}{\sqrt{B_i^2+C_i^2}}\cos\omega t+\frac{C_i}{\sqrt{B_i^2+C_i^2}}\sin\omega t\bigr)=$

$=A_i\cos(\omega t+\omega_i)$ , where $A_i=\frac{1}{\sqrt{B_i^2+C_i^2}}$ and $\omega_i$ such number that $\cos\omega_i=\frac{B_i}{\sqrt{B_i^2+C_i^2}}$ and $\sin\omega_i=-\frac{C_i}{\sqrt{B_i^2+C_i^2}}$

We obtain $\vec{r}(t)=(A_1\cos(\omega t+\omega_1),\ldots,A_n\cos(\omega t+\omega_n))$.

Initial condition gives us $(x_0,0,\ldots,0)=\vec{r}(0)=(A_1\cos\omega_1,\ldots,A_n\cos\omega_n)$, so $A_1\cos\omega_1=x_0$ and $A_i\cos\omega_i=0$ while $i>1$.

This means that 1)$A_1\cos\omega_1=x_0$ and 2)$A_i=0$ or $\cos\omega_i=0$ for all $i>1$.

Anyway, it was obtained that $\vec{r}(t)=(A_1\cos(\omega t+\omega_1),\ldots,A_n\cos(\omega t+\omega_n))$ is an equation of simple harmonic motion.