# Central Hooke's Law Force

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1. Oct 8, 2014

### pixatlazaki

1. The problem statement, all variables and given/known data
A particle of mass $m$ is placed on a smooth table and attached to a fixed point $O$ on the table by a spring with spring constant $k$ and natural length $l$.

(i) Show that the particle can execute circular motion about $O$ with angular velocity $\omega$ provided $\omega < k/m$ , and calculate the resulting extension $e$ of the spring.

(ii) Show that the frequency $\Omega$ of small oscillations of the particle about the circular motion computed in (i) is given by $$\Omega = \omega\sqrt{4 + \frac{l}{e}}$$

2. Relevant equations
$$F_s = -ke = -k(r-l) = -\frac{dU}{dr}$$
$$m\ddot{r} = mr\omega^2 - \frac{dU}{dr}$$ (from Newton's 2nd or the Euler-Lagrange equation)
$$r-l=e$$

3. The attempt at a solution
(i). In this case, $\ddot{r}=0$, as $r=$const. Thus,
$$mr\omega^2 - k(r-l) = 0$$
$$kl = kr - mr\omega^2$$
$$\frac{k}{m}l = \frac{k}{m}r - \omega^2r$$
$$\frac{k}{m}l= (\frac{k}{m} - \omega^2)r$$
$r>0$ and $\frac{k}{m}l > 0$ (as $k,m,l > 0$) so for this equation to hold (and therefore, for $\ddot{r}$ to equal $0$, $(\frac{k}{m} - \omega^2)>0$, so $\omega<\sqrt{k/m}$.
Plugging in $r=l+e$.
$$\frac{k}{m}l = (\frac{k}{m} - \omega^2)(l+e)$$
$$\frac{k}{m}l = (\frac{k}{m} - \omega^2)l + (\frac{k}{m} - \omega^2)e$$
$$(\ast)\:\frac{\omega^2l}{\frac{k}{m}-\omega^2} = e$$

(ii). The radial equation of motion (plugging in $r=l+e$ once again) is
$$m(l+e)\omega^2 - ke = m\ddot{e}$$
$$m\omega^2l + m\omega^2e - ke = m\ddot{e}$$
$$\ddot{e} + (\frac{k}{m} - \omega^2)e = \omega^2l$$
$$e(t) = e_{hom}(t) + e_{part}(t)$$
$$e(t) = A\cos((\frac{k}{m} - \omega^2)t-\delta) + \frac{\omega^2l}{\frac{k}{m} - \omega^2}$$, where $A, \delta$ are some undetermined constants from the solving of the differential equation.
The (angular) frequency of $e(t)$ is then $$\Omega = \sqrt{\frac{k}{m} - \omega^2}$$
Solving $(\ast)$ for $\frac{k}{m} - \omega^2$, we find
$$\frac{\omega^2l}{e} = \frac{k}{m} - \omega^2$$
Thus,
$$\Omega = \omega\sqrt{\frac{l}{e}}$$

Apologies for the long post. Where exactly am I going wrong?

2. Oct 9, 2014

### Staff: Mentor

I feel that you also need to consider the force balance in the theta direction, since, with the oscillatory motion, $\omega$ is varying with r.

Chet