Harmonic motion

[Sriman Dutta]

Please check whether my solution to the problem is right or not.

[Sriman Dutta]

Can someone help?

[studiot]

Can you please post your pics the right way up?

 

It is a nuisance to turn them round.

 

You have not made it clear why there should be any oscillation at all.

In other words why will the spring not simply stretch until there is a balance between the electrostatic attraction and the spring force?

 

Note I am not saying there will be no oscillation, just that you haven't established the conditions for it to occur.

[Bender]

A typed explanation of the problem and clearer pics are going to increase the odds of someone responding.

[Sriman Dutta]

Well I didn't turn them..... They were erect. But when I uploaded the pics they turned about.

 

Now coming to the problem.

 

A charged ball with charge q is attached to a spring of stiffness k. At some distance there is a charged plate, of charge Q. So as soon as the electrostatic force (q and Q are opposite) exerts a pull on the ball towards the plate, a restoring force builds up in the spring in the opposite direction. So the equation becomes

 

[math] m\frac{d^2x}{dt^2}=-kx+F [/math] , F is the electrostatic force and m is the mass of the ball. x is the displacement of the ball about its mean position.

But as the ball shows a harmonic motion, the external electrostatic force acting upon it is also harmonic or periodic in nature. Considering that F shows oscillations, the equation becomes

 

[math]m\frac{d^2x}{dt^2}=-kx+\frac{k_e qQ}{d_0^2}sin \omega t [/math]

Solving this yields [math]x(t)= \frac{k_e qQ}{d_0^2 (m\omega^2+k)}sin \omega t [/math]

Here ke is Coulombs constant and [math]\omega[/math] is the angular frequency, experimentally determined.

 

Am I correct??

[Bender]

The Coulomb force should depend on the displacement x. I don't think you can assume it will be sinusoidal in time.

I think you also used the equation for the force between two spherical charges, rather than between sphere and plate.

[studiot]

Well I didn't turn them..... They were erect. But when I uploaded the pics they turned about.

 

Now coming to the problem.

 

A charged ball with charge q is attached to a spring of stiffness k. At some distance there is a charged plate, of charge Q. So as soon as the electrostatic force (q and Q are opposite) exerts a pull on the ball towards the plate, a restoring force builds up in the spring in the opposite direction. So the equation becomes

 

[math] m\frac{d^2x}{dt^2}=-kx+F [/math] , F is the electrostatic force and m is the mass of the ball. x is the displacement of the ball about its mean position.

But as the ball shows a harmonic motion, the external electrostatic force acting upon it is also harmonic or periodic in nature. Considering that F shows oscillations, the equation becomes

 

[math]m\frac{d^2x}{dt^2}=-kx+\frac{k_e qQ}{d_0^2}sin \omega t [/math]

Solving this yields [math]x(t)= \frac{k_e qQ}{d_0^2 (m\omega^2+k)}sin \omega t [/math]

Here ke is Coulombs constant and [math]\omega[/math] is the angular frequency, experimentally determined.

 

Am I correct??

 

I think you are trying to make this too complicated and should go back to the basics of SHM. You have too many terms in your equations for SHM.

Further you have not responded to my hints in post#3. Your equations are overcomplicated because you conditions are inappropriate. You have yet to place the mass at the beginning of then process or establish what is meant by 'the mean position'.

 

Bender asked you to state the original question you are trying to solve as written. Please do so as I wonder if you are not missing something out?

 

 

 

Quite separately, I have started a thread in Feedback to try to get some help with placing images here.

Edited March 30, 2017 by studiot

[Sriman Dutta]

Firstly the electric field will always point towards the plate, that is, the direction of electrostatic force remains unchanged. The restoring force acts in opposite directions with respect to the displacement x. This means that if the electrostatic force attracts the ball due to opposite charges, a positive displacement towards the plate occurs. However a restoring force acts in the opposite direction. The displacement towards the plate occurs as long as F>kx. But at the extreme position, when kx> F, the ball comes back. This continues and causes a negative displacement.

This makes me think that the ball shows harmonic motion.

Yes I doubt whether the F is sinusoidal or not. But magnitude of F is dependent on x, and if we write it like this

F=kqQ/(l-x)^2, where l is distance of plate from mean position.

But if I put this value in the equation, it becomes so tedious. Therefore I assume that F=kqQ/d^2 sin wt

[Bender]

Try a coordinate transformation so that

[math]F=\frac{kqQ}{(x')^2}[/math]

 

Not sure whether that will be easily solvable, but another substitution might do it. How familiar are you with solving differential equations?

 

If you are new to this kind of thing, it is probably better to start with a simpler problem that doesn't involve an inverse quadratic dependency.

Edited March 30, 2017 by Bender

[John Cuthber]

To get simple harmonic motion the restoring force has to be proportional to the displacement.

That's probably true (or near enough) for the spring.

But the force between two conductors is not proportional to the distance between them.

So the motion will not be simple harmonic.

 

Also, if you are talking about this sort of thing

https://en.wikipedia.org/wiki/Oxford_Electric_Bell

then the motion is not SHM, it changes direction suddenly when the ball hits the bell.

[zztop]

Firstly the electric field will always point towards the plate, that is, the direction of electrostatic force remains unchanged. The restoring force acts in opposite directions with respect to the displacement x. This means that if the electrostatic force attracts the ball due to opposite charges, a positive displacement towards the plate occurs. However a restoring force acts in the opposite direction. The displacement towards the plate occurs as long as F>kx. But at the extreme position, when kx> F, the ball comes back. This continues and causes a negative displacement.

This makes me think that the ball shows harmonic motion.

Yes I doubt whether the F is sinusoidal or not. But magnitude of F is dependent on x, and if we write it like this

F=kqQ/(l-x)^2, where l is distance of plate from mean position.

But if I put this value in the equation, it becomes so tedious. Therefore I assume that F=kqQ/d^2 sin wt

The correct equation is:

 

[math]m \frac{d^2x}{dt^2}+kx+\frac{a}{x^2}=0[/math]

 

The solution is:

 

this

 

The motion is not harmonic

Edited March 30, 2017 by zztop

[Sriman Dutta]

Here's a rough illustrations.....

I want to know why you think that it isn't harmonic? I thought it as an example of forced vibration.

[Bender]

It is not a forced vibration, because there is no externally imposed frequency. It is free to vibrate at the natural frequency of the system (given that it is not harmonic, the frequency probably depends on the amplitude).

 

It is not harmonic because the ball is not going to move sinusoidally, for the reasons given by John Cuthber.

[studiot]

Upon rereading all the posts and the title, I owe you an apology, Sriman.

 

Nowhere have you mentioned simple harmonic motion, just harmonic motion which is different.

 

However as I already indicated you have not convinced me that the mass will move at all, let alone oscillate.

 

You should start by considering the Physics, before throwing in equations of any sort.

I think this applies to some other posts as well.

 

For example the nature of charged plate is important, certainly the distance dependence of the electric force is not as anyone has as yet stated.

 

If the plate is large enough to be considered 'infinite', then the force is independent of distance, ie it is a constant.

 

If the plate is of finite dimensions then at a large enough distance from it the attraction does indeed approach Qq/z², but the true relationship is subject to

 

[math]\left[ {\frac{z}{{\sqrt {\left( {{a^2} + {z^2}} \right)} }} - 1} \right][/math]

Where z is the distance from the plate and a is the radius of a circular plate.

Then we must consider the spring.

Is the spring able to be compressed or not?

If so then the spring force may reverse in direction at some point in any motion.

Finally there is the starting position.

Are we starting from an unstretched spring and, holding the mass in position and then releasing it, or somehow suddenly switching on the electricity?

Or are we starting from a position (if one exists) where the restraining force of the spring exactly equals the electric force of attraction.

If so why will the mass move at all?

It is clear that there could be different equations in operation at different points along the axis connecting the spring foundation and the charged plate and that some of them may depend upon the dimensions of the system.

So this is the last time I will ask you to specify properly.

Edited March 31, 2017 by studiot

[Sriman Dutta]

Initially the plate is uncharged. Then at that point the ball lies in its origin or original position or initial position. There no forces are acting. And I am considering that initial position as the reference . Displacement to the left of the initial position is considered as positive and on the other side it's negative. Then I introduce the charge on the plate. Due to this the ball will show a movement.

[studiot]

Initially the plate is uncharged. Then at that point the ball lies in its origin or original position or initial position. There no forces are acting. And I am considering that initial position as the reference . Displacement to the left of the initial position is considered as positive and on the other side it's negative. Then I introduce the charge on the plate. Due to this the ball will show a movement.

 

Yes I know the mass starts off in its initial position. That is a definition of initial position.

 

But where is this initial position?

 

And what of the spring? What is its condition?

 

Come on, all the equations on earth are useless without the proper boundary conditions as well.

 

I suggest you leave the location of the zero point (origin) until after the boundary conditions are fully specified, there is much to be said for measuring from the charged plate.

 

Usually when analysing a system either

 

The system is in a neutral condition in which case an outside agent acting once only is invoked to provide an initial displacement and the system then follows its trajectory controlled by the system forces.

 

or

 

The system is held at some point in the trajectory and released at time t = 0 and the system again follows its trajectory.

 

A linear system will repeat the same trajectory over and over.

 

But this system is non linear and I think there is the possibility that the neutral point will be displaced along the horizontal axis with each cycleand that each cyle will be different.

 

So what I think you are saying is.

 

The mass is attached to the unstretched spring (although you have not said if the spring can work in compression.

 

Before t=0 no forces are acting

 

The charge is somehow 'switched on' at t=0.

 

The electric attractive force E causes the mass to move towards the plate, stretching the spring which provides a restraining force S.

 

Now E increases as a second order quantity of displacement, whilst S increases as first order.

So even if E starts from a lower base than S (depending upon the constants involved in the boundary conditions) it will eventually overtake S and from that point S will not be able to return the mass towards its original position.

Edited March 31, 2017 by studiot

[studiot]

Does this question have anything to do with the original?

 

[Sriman Dutta]

I am not sure...........but I think that (A) is the correct answer.

[Roger Dynamic Motion]

SHM ~ The both ends of the Oscillator, must move ~ meaning; change positions while oscillating ! EXAMPLE ...

 

click on (https)

.Retro causality phenomenon:

Model shown wire frame. Expansion of Euler rotations; Precession , Rotation , Nutation. Since the mathematics is somewhat complicated, you should read the PD F file on the Gyroscope Precession and Nutation for detailed derivations of these results. The axis of rotation of the body frame, precession axis, is connected to the top with the world, blue frame . All the blue parts are fixed together and move relative to each other . The other frame, the green frame, is called the body frame, representing the rotation around itself,= the moving frame.Here is some of the functions that govern the motion of the systems.The body frame,~ the green frame, contains a MASS at the end of the axis that weight X pounds and that axis is rigidly connected to the precession axis at the bottom and rotate at it's base around the x axis of the green plate (frame) witch frame is free to rotate around the world blue axis as seen from the model. Note that the two frames rotate independently around the same blue axis which is the world axis. Now lets work with gravity. The orange body in the model is raised against gravity with an acceleration 64 FEET (S) by the rotation of the two frames; rotation opposite to each other, with the help of the oscillator bringing the two knobs (the green square and the blue square) of the two frames closer to each other by shrinking time as needed witch is calibrated..til the right amount of acceleration necessary to raise the weight, at 64 feet second is reached beating gravity from so doing.The restoring force is generated by the gravitational forces when the orange body is at is highest and furthers point (position)... from the center of the system and the oscillator at that time is at zero Potential ;...but the mass (weight) is at it's maximum and is forced back down by the G.F. to its original position,from the start so! by so doing, restoring the potential of the oscillator...(stretching the spring ). That's what makes this concept great. It beets gravity and again, gravity is the restoring force from stretching the oscillator (spring)= ( the potential of the Oscillator) . If you observe at the oscillator, the two FRAMES, work together to raise the weight. When the oscillator forces the green frame to rotate to the right and at the same time the blue frame rotates to the left . The blue frame transfers its motion to the top through the world AXIS (the blue axis) holding the precession AXIS while the green ( frame), rotates to the right raising the mass; so they both conspire to raise the orange body while the green ( frame) is rotating the axis that carry the mass at the same time. A load, torque clockwise is applied to the system which slows it down. see model 65

an error has appended. Sorry about that .../

[Roger Dynamic Motion]

The S~ harmonic motion of an oscillator; must have a complement of reaction.

[Roger Dynamic Motion]

I am not sure...........but I think that (A) is the correct answer.__/ I believe So!

[Roger Dynamic Motion]

S H M ~ can be understood like action & reaction .