# 2nd order differential equation - equation of motion

• MHB
• jd17
In summary: If I start with the 40x acting to the right, then the solving the differential equation does not end up as a harmonic equation.If you could find a way to explain this situation to a 'bear of very little brain' I would be grateful.
jd17
Hi
There is an example in my textbook worded as follows;

A particle of mass 2kg moves along the positive x-axis under the action of a force directed towards the origin. At time t seconds, the displacement of P from O is x metres and P is moving away from O with a speed of v ms^-1. The force has magnitude 40x N. The particle P is also subject to resistive forces of magnitude 16v N.

a) Show that the equation of motion of P is d^2x/dt^2 + 8 dx/dt +20x = 0

I do not understand how the force can be towards the origin and the particle be moving away from the origin?
The diagram in the example shows the origin O to the left of P with the force 40x pointing towards the left , the origin, as well as the resistive force 16v and the acceleration pointing to the right.

I do not understand how you can draw the force of 40x and the resistive force of 16v pointing in the same direction.
If P is moving away from the origin then should the force of 40x be directed to the right, away from O.

However in the algebra they use F=ma with -40x - 16v =2a
Whilst I understand that the -40x implies the force acting in the opposite direction to what is shown in the diagram and therefore to the right and being opposed by the resistive force, this seems to be a very contrived why to end up with a damped harmonic motion when the differential equation is solved.
{given x(0)=0 and x(pi/4)=3e^-pi}

If I start with the 40x acting to the right, then the solving the differential equation does not end up as a harmonic equation.

If you could find a way to explain this situation to a 'bear of very little brain' I would be grateful.

I have tried to think of a practical context where this force and resistance might occur and am thinking of a car accelerating from O at a uniform rate - the 'force' on the accelerator increasing uniformly as x increases and the resitive force being the air resistance which increases with velocity, but I can't see how air resistance, being a multiple of v, would stop the car. Do you know of any practical situation that this model would represent?

Thank you in advance for any assistance you can give

jd17 said:
Hi
There is an example in my textbook worded as follows;

A particle of mass 2kg moves along the positive x-axis under the action of a force directed towards the origin. At time t seconds, the displacement of P from O is x metres and P is moving away from O with a speed of v ms^-1. The force has magnitude 40x N. The particle P is also subject to resistive forces of magnitude 16v N.

a) Show that the equation of motion of P is d^2x/dt^2 + 8 dx/dt +20x = 0

I do not understand how the force can be towards the origin and the particle be moving away from the origin?
Haven't you ever walked into a strong wind? The wind is applying a force on you that is opposite to the way you are walking. Of course there has to be an additional force, due to your feet, or some initial velocity. "Force= mass times acceleration" so the acceleration is in the direction of the (net)
force. Velocity does not have to be.

The diagram in the example shows the origin O to the left of P with the force 40x pointing towards the left , the origin, as well as the resistive force 16v and the acceleration pointing to the right.

I do not understand how you can draw the force of 40x and the resistive force of 16v pointing in the same direction.
If P is moving away from the origin then should the force of 40x be directed to the right, away from O.
Not necessarily, the force, 40x, is 40 times the x coordinate. That has nothing to do with the velocity so nothing to do with "P is moving away from the origin". More importantly, the problem specifically says "a force directed towards the origin". As long as x is positive, so the particle is to the right of the origin, the force is directed back to the left. The problem also says "The force has magnitude 40x N". Notice the word "magnitude". That is the "strength" of the force and says nothing about the direction. As long as x is positive, that force vector is -40x because it is directed back toward the origin.

However in the algebra they use F=ma with -40x - 16v =2a
Whilst I understand that the -40x implies the force acting in the opposite direction to what is shown in the diagram and therefore to the right
?? No, as long as x is positive (to the right), -40x is negative so back to the left.

and being opposed by the resistive force, this seems to be a very contrived why to end up with a damped harmonic motion when the differential equation is solved.
I'm not sure what you mean by this. Any "resistive force", pretty much by definition, "resists" the motion. A force in the same direction as the velocity would increase it, not "resist" it.
{given x(0)=0 and x(pi/4)=3e^-pi}

If I start with the 40x acting to the right, then the solving the differential equation does not end up as a harmonic equation.
Yes, if the force is in the same direction as the motion, the object will just keep going in the same direction.

If you could find a way to explain this situation to a 'bear of very little brain' I would be grateful.

I have tried to think of a practical context where this force and resistance might occur and am thinking of a car accelerating from O at a uniform rate - the 'force' on the accelerator increasing uniformly as x increases and the resitive force being the air resistance which increases with velocity, but I can't see how air resistance, being a multiple of v, would stop the car. Do you know of any practical situation that this model would represent?
If are driving along a road and take your foot off the accelerator but don't press on the brake pedal, the car will slow and eventually stop because of resistance (more friction in the car and tires than air).

Thank you in advance for any assistance you can give
"A particle of mass 2kg moves along the positive x-axis under the action of a force directed towards the origin. At time t seconds, the displacement of P from O is x metres and P is moving away from O with a speed of v ms^-1. The force has magnitude 40x N. The particle P is also subject to resistive forces of magnitude 16v N."
The force is "directed towards the origin" and "has magnitude 40x" so, as a one dimensional vector, is -40x. There is also a resistive force of magnitude 16v. Again, that is magnitude and says nothing about the direction. The fact that it is "resistive" means that it is opposite the direction of v: -16v. The total force is -40x- 16v so "force equals mass times acceleration" gives 2a= -40x- 16v. The velocity is the derivative of the position vector and acceleration is the derivative of v so the second derivative of x. We can write that equation as 2x''= -40x- 16x' or 2x''+ 16x+ 40x= 0. That is a second order equation with constant coefficients. It has "characteristic equation" 2r^2+ 16r+ 40= 0. That is equivalent to r^2+ 8r+ 20= r^2+ 8r+ 16+ 4= (r+ 4)^2+ 4= 0. That has solutions r= -4- 2i and r= -4+ 2i. The general solution to the differential equation is $$x(t)= e^{-4t}(Acos(2t)+ Bsin(2t))$$.

Last edited:
Country Boy said:
Haven't you ever walked into a strong wind? The wind is applying a force on you that is opposite to the way you are walking. Of course there has to be an additional force, due to your feet, or some initial velocity. "Force= mass times acceleration" so the acceleration is in the direction of the (net)
force. Velocity does not have to be.

This is where I am still confused. for me to move to the right, against a wind blowing to the left I need to apply a driving force, otherwise I will remain still or move backwards if the wind is strong enough.
The only 'driving force' mentioned in the question is 'directed towards the origin, so how can the object be moving away from the origin

Country Boy said:
Not necessarily, the force, 40x, is 40 times the x coordinate. That has nothing to do with the velocity so nothing to do with "P is moving away from the origin".

If the 40x is the only force that drives the movement of the object then why would this force have nothing to do with the velocity?
Country Boy said:
More importantly, the problem specifically says "a force directed towards the origin". As long as x is positive, so the particle is to the right of the origin, the force is directed back to the left.

Still confused, how can the only driving force be directed towards the left but the object be moving to the right unless there was an an initial velocity, but there is no mention of that in the question?

Country Boy said:
The problem also says "The force has magnitude 40x N". Notice the word "magnitude". That is the "strength" of the force and says nothing about the direction. As long as x is positive, that force vector is -40x because it is directed back toward the origin.
Country Boy said:
?? No, as long as x is positive (to the right), -40x is negative so back to the left.

I'm not sure what you mean by this. Any "resistive force", pretty much by definition, "resists" the motion. A force in the same direction as the velocity would increase it, not "resist" it.

Yes, if the force is in the same direction as the motion, the object will just keep going in the same direction.

If are driving along a road and take your foot off the accelerator but don't press on the brake pedal, the car will slow and eventually stop because of resistance (more friction in the car and tires than air).

I understand this system, as there is an initial velocity to the right so once the foot is removed from the accelerator the car will slow down due to the various resistive forces, unless it was going down a hill...

Thank you for the comments you have made but I still seem to be missing some basic understanding here
John

Country Boy said:
"A particle of mass 2kg moves along the positive x-axis under the action of a force directed towards the origin. At time t seconds, the displacement of P from O is x metres and P is moving away from O with a speed of v ms^-1. The force has magnitude 40x N. The particle P is also subject to resistive forces of magnitude 16v N."
The force is "directed towards the origin" and "has magnitude 40x" so, as a one dimensional vector, is -40x. There is also a resistive force of magnitude 16v. Again, that is magnitude and says nothing about the direction. The fact that it is "resistive" means that it is opposite the direction of v: -16v. The total force is -40x- 16v so "force equals mass times acceleration" gives 2a= -40x- 16v. The velocity is the derivative of the position vector and acceleration is the derivative of v so the second derivative of x. We can write that equation as 2x''= -40x- 16x' or 2x''+ 16x+ 40x= 0. That is a second order equation with constant coefficients. It has "characteristic equation" 2r^2+ 16r+ 40= 0. That is equivalent to r^2+ 8r+ 20= r^2+ 8r+ 16+ 4= (r+ 4)^2+ 4= 0. That has solutions r= -4- 2i and r= -4+ 2i. The general solution to the differential equation is $$x(t)= e^{-4t}(Acos(2t)+ Bsin(2t))$$.

jd17 said:
This is where I am still confused. for me to move to the right, against a wind blowing to the left I need to apply a driving force, otherwise I will remain still or move backwards if the wind is strong enough.
The only 'driving force' mentioned in the question is 'directed towards the origin, so how can the object be moving away from the origin
If the 40x is the only force that drives the movement of the object then why would this force have nothing to do with the velocity?

Still confused, how can the only driving force be directed towards the left but the object be moving to the right unless there was an an initial velocity, but there is no mention of that in the question?
The problem specifically said that the object was moving to the right. There must have been some "initial velocity", to the right.
tand this system, as there is an initial velocity to the right so once the foot is removed from the accelerator the car will slow down due to the various resistive forces, unless it was going down a hill...

Thank you for the comments you have made but I still seem to be missing some basic understanding here
John

## 1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves a function and its first and second derivatives. It is commonly used to describe the motion of a physical system, such as a particle or a spring.

## 2. What is the equation of motion for a 2nd order differential equation?

The equation of motion for a 2nd order differential equation is typically written in the form of ma = F, where m is the mass of the object, a is its acceleration, and F is the net force acting on the object. This equation can be derived from Newton's second law of motion.

## 3. How is a 2nd order differential equation solved?

There are various methods for solving a 2nd order differential equation, depending on the specific equation and initial conditions. Some common methods include separation of variables, substitution, and using power series or Laplace transforms.

## 4. What are the applications of 2nd order differential equations in physics?

2nd order differential equations are used to model a wide range of physical systems, including the motion of objects under the influence of gravity, simple harmonic motion, and electrical circuits. They are also used in fields such as mechanics, thermodynamics, and quantum mechanics.

## 5. How do 2nd order differential equations relate to real-world problems?

2nd order differential equations are essential for understanding and predicting the behavior of physical systems in the real world. They allow scientists and engineers to analyze and design systems, such as bridges, airplanes, and electronic devices, by describing their motion and behavior mathematically.

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