Finding the Position(s) of Particle Oscillation

[Ajay N]

Homework Statement

A particle of mass m is connected from the origin O of an inertial frame with the help of an elastic cord of force constant k and negligible relaxed length. In addition to the elastic force of the cord, a constant force F=F î also acts on the particle.The particle is pulled to an arbitrary position (x0,y0) as in the figure and released.Find the position between which the particle oscillates and period of the oscillations.

Relevant Equations

a=-(ω^2)x
T=2π/ω

My first impression was that since the external force was constant it wouldn't make any change on the period of oscillation of the system. But on further thinking I found that if I were to only consider the one dimensional oscillation of the particle then the component of force along the string will not remain a constant and will change depending on its position. And then I got stuck on it. Plus I don't have any idea on the position between which the particle oscillates.

 

[mfb]

Did you find the equation of motion?
A suitable coordinate transformation should get rid of the external force.

 

[Ajay N]

Well I don't think I know how to transform the coordinates. I might have to look into that.

 

[kuruman]

Well I don't think I know how to transform the coordinates. I might have to look into that.

If you first find the equation of motion as @mfb suggested, it will be easier to figure out the suitable coordinate transformation.

 

[mfb]

You don't need to solve it via a coordinate transformation, that is just the easiest approach.

But what you absolutely need is writing down the equation of motion, because complaining that you don't know how to do the third step already while not doing the first step isn't getting you anywhere.

 

[rude man]

You might take note of the fact that this problem is equivalent to the same cord hanging vertically. The force F is then gravity. It's somewhat weirdly described but that is my interpretation.

On top of that, in polar coordinates and even assuming $θ_0$ = arc tan (y0/x0) << 1 I get a system of non-linear ODE's in r and θ, r = +√$(x^2 + y^2)$. So, looks difficult, at least to me.

 

[kuruman]

You might take note of the fact that this problem is equivalent to the same cord hanging vertically. The force F is then gravity. It's somewhat weirdly described but that is my interpretation.

On top of that, in polar coordinates and even assuming $θ_0$ = arc tan (y0/x0) << 1 I get a system of non-linear ODE's in r and θ, r = +√$(x^2 + y^2)$. So, looks difficult, at least to me.

My interpretation is that there is no "vertical". The mass is on a horizontal frictionless plane.

 

[rude man]

My interpretation is that there is no "vertical". The mass is on a horizontal frictionless plane.

I said "equivalent". The force depicted here as horizontal is constant so you can just flip the diagram 90 degrees. That force now becomes gravity.
Not that that helps much in solving. I myself couldn't handle the ODE's describing this highly coupled, non-linear system of equations..

 

[mfb]

@rude man: Try Cartesian coordinates.

 

[rude man]

@rude man: Try Cartesian coordinates.

I started to, but that looked even worse. Is that your recommendation?
Generalized coordinates (Lagrangians)?

 

[Ajay N]

Can I take the system as two independent oscillators oscillating along x and y-axis ? By this method the oscillation along the y-axis is completely unaffected by the external force. The force along the x-axis being a constant doesn't affect the time period of the oscillation along x axis. From this I get the answer as
T=2π/ω where ω=√k/m

The above answer is given as the solution to this problem in my book.

For the coordinate between which the oscillation takes place, it can be found out using conservation of energy.

 

[PeroK]

Can I take the system as two independent oscillators oscillating along x and y-axis ? By this method the oscillation along the y-axis is completely unaffected by the external force. The force along the x-axis being a constant doesn't affect the time period of the oscillation along x axis. From this I get the answer as
T=2π/ω where ω=√k/m

The above answer is given as the solution to this problem in my book.

For the coordinate between which the oscillation takes place, it can be found out using conservation of energy.

You need some justification to treat the oscillations as independent.

Did you ever get round to deriving the equations of motion?

 

[Ajay N]

I tried deriving the equation but with no success. I am a high school student. Can the equation of motion be derived with my limited mathematical knowledge ? But this problem is from a high school physics olympiad book.

 

[PeroK]

I tried deriving the equation but with no success. I am a high school student. Can the equation of motion be derived with my limited mathematical knowledge ? But this problem is from a high school physics olympiad book.

Show us how much you can do. What if we ignore $F$ for the moment? Could you derive the equations of motion in that simplified case?

 

[Ajay N]

In that case x=x₀cos(ωt) and y=y₀cos(ωt).

 

[PeroK]

In that case x=x₀cos(ωt) and y=y₀cos(ωt).

But, how did you derive those? If you simply quote them, the problem is what do you do if something is slightly different. You need to understand where those solutions come from.

 

[Ajay N]

The net force acting on the particle at (x,y) is k√(x²+y²) .
F=k√(x²+y²)
The component of this force along x-axis is F_x
F_x=k√(x²+y²) * x/√(x²+y²)=kx.
From this we can infer that the x coordinate is oscillating. Amplitude is x₀. At t=0 x=x₀ therefore x=x₀cos(ωt).
Similarly we can derive y₀cos(ωt).

 

[PeroK]

The net force acting on the particle at (x,y) is k√(x²+y²) .
F=k√(x²+y²)
The component of this force along x-axis is F_x
F_x=k√(x²+y²) * x/√(x²+y²)=kx.
From this we can infer that the x coordinate is oscillating. Amplitude is x₀. At t=0 x=x₀ therefore x=x₀cos(ωt).
Similarly we can derive y₀cos(ωt).

What if you add the constant force $F$ to those equations?

 

[Ajay N]

Then there would be no change in the the time period since F is a constant.

 

[PeroK]

Then there would be no change in the the time period since F is a constant.

That's not a mathematical analysis; and, it doesn't produce the full solution. Just add it to your equations.

 

[Ajay N]

F_y=-ky
F_x=-kx+F

 

[kuruman]

F_y=-ky
F_x=-kx+F

Can you write Newton's second law based on these equations? These would be the equations of motion.

 

[Ajay N]

ma_x=-kx+F or m d²x/dt²=-kx+F
ma_y=-ky or m d²y/dt²=-ky

 

[rude man]

ma_x=-kx+F or m d²x/dt²=-kx+F
ma_y=-ky or m d²y/dt²=-ky

This is correct. Once you take an elementary differential equations course you'll realize these are extremely simple to solve without worrying about any "coordinate transformations". I must admit that without this background it's perhaps too involved for a high school student. In any case I wouldn't have solved it without that background.

What threw me at first is the requirement that the spring have zero relaxed length. Otherwise it could not have been solved, at least not readily.

 

[kuruman]

This is correct. Once you take an elementary differential equations course you'll realize these are extremely simple to solve without worrying about any "coordinate transformations". I must admit that without this background it's perhaps too involved for a high school student. In any case I wouldn't have solved it without that background.

What threw me at first is the requirement that the spring have zero relaxed length. Otherwise it could not have been solved, at least not readily.

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.

 

[mfb]

The particle oscillates in a straight line in this plane, with well-defined end positions with zero velocity.

 

[rude man]

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.

What's "FBD"?
Computing $r_{max}$ and $r_{min}$ does not by itself constitute "the position between which the particle oscillates ..., does it? You'd also have to give the respective angles, wouldn't you?.
I'd stick with solving for x(t) and y(t) for all t. Easily done but you do need some elementary ODE knowledge.

 

[haruspex]

Well I don't think I know how to transform the coordinates. I might have to look into that.

It's quite easy. Just substitute x=x'+c, where c is a constant, in your equation and look for a value of c that makes it simpler.

 

[kuruman]

What's "FBD"?
Computing $r_{max}$ and $r_{min}$ does not by itself constitute "the position between which the particle oscillates ..., does it? You'd also have to give the respective angles, wouldn't you?.
I'd stick with solving for x(t) and y(t) for all t. Easily done but you do need some elementary ODE knowledge.

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.

 

[rude man]

One can show by parametrization (eliminating t) that the particle oscillates in a straight line in x-y space. This parametrization with t gives a complete picture of the particle's locus and confirms the observation of post 26.

 

[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.