Finding the Position(s) of Particle Oscillation

[rude man]

FBD stands for "Free Body Diagram", the acronym is widely used in PF. If one draws a FBD, it is easy to see that there is an equilibrium point at which the net force is zero and with respect to which the net torque is also zero. With respect to that point, angular momentum is conserved and the mass executes one dimensional harmonic motion. The amplitude can be found by energy conservation without solving an ODE. My question was how to interpret "the position between which the particle oscillates". It seems that two position vectors relative to the given origin are required to answer the question and we already know one of them.

From this viewpoint I would say the position vectors are (0,0) and ($x_0$, $y_0$) since the locus is of the form $x = k_1y + k_2$
where $k_1$ is fully described by $x_0$ and $y_0$ and $k_2$ is a constant independent of $x_0$ or $y_0$. Cf. my post # 30.

 

[SammyS]

Yet if one looks at the figure in post #1 one must agree that there is a torque about point O at the time of release. Thus there is need of a third equation linking the torque to the angular acceleration. The question then is how to interpret the statement "Find the position between which the particle oscillates and period of the oscillations" when the motion takes place in a plane. It is possible to find $r_{max}$ and $r_{min}$ from the equilibrium point $[F/k, 0]$ using energy conservation and a FBD when the radial component of the velocity is zero.

It is interesting to note that this is the first post in the thread to mention that the stated question includes the task of finding the position(s) between which the particle oscillates.

With the equilibrium position of the system being the point $\ \left(\dfrac{F}{k} , 0 \right)\,,\$ the remaining position is easy to find, assuming that the particle is released from rest at position $\left(x_0 , y_0 \right)$ .

 

[rude man]

It is interesting to note that this is the first post in the thread to mention that the stated question includes the task of finding the position(s) between which the particle oscillates.

With the equilibrium position of the system being the point $\ \left(\dfrac{F}{k} , 0 \right)\,,\$ the remaining position is easy to find, assuming that the particle is released from rest at position $\left(x_0 , y_0 \right)$ .

Quite so. Since only the extrema were asked for.
We can derive two equations:
one, conservation of energy between the two extrema (x0,y0) and (x1,y1).
two, net forces along x = 0 at (x1,y1).
That does neatly avoid having to solve the ODEs.