Classical mechanics differential equation F(x) = -kx

[Salvador_]

Homework Statement

A particle of mass m is subject to a force F (x) = -kx. The initial position is
zero, and the initial speed is v₀. Find x(t).

Homework Equations

F = m*v*dv/dx = -kx
v = dx/dt

The Attempt at a Solution

I'm new to differential equations, so please excuse me if I make any amateurish mistakes.
-kx = mv*dv/dx
After integrating:
-kx²/2 + c = mv²/2
v² = -kx² + c

This is the part I'm unsure about, the inital values:
v²(0) = 0 + c = v₀²
So c = v₀²

v = (v₀² - kx²/m)^1/2

dx/v(x) = dt
After integrating: (a bit messy)
t = (m/k)^1/2*arcsin(x*(k/v₀²m)^1/2) + c
c=0
Just wanted to know if I went wrong somewhere as I'm not convinced by the final answer.

 

[Orodruin]

You are to find x as a function of t, it makes more sense to solve the differential equation in time. You can do it like that if you really want to complicate your life... just solve for x from your expression.

 

[Salvador_]

You are to find x as a function of t, it makes more sense to solve the differential equation in time. You can do it like that if you really want to complicate your life... just solve for x from your expression.

How would I go about doing this? I'm not sure how to solve in time if the function is of position.

 

[Orodruin]

How would I go about doing this? I'm not sure how to solve in time if the function is of position.

What function is of position? There are only functions of time. The force is a function of position, which is a function of time. What you have to do is to solve the resulting differential equation.

 

[Orodruin]

Also, the work-energy theorem is not really what you want to use. Newton 2 will suffice.

 

[Salvador_]

What function is of position? There are only functions of time. The force is a function of position, which is a function of time. What you have to do is to solve the resulting differential equation.

Then I'd have dv/dt = -kx
dv = -kx*dt
I'm guessing that's not what you meant because the integral doesn't make sense.

 

[Orodruin]

Then I'd have dv/dt = -kx
dv = -kx*dt
I'm guessing that's not what you meant because the integral doesn't make sense.

Of course it makes sense (the first expression is what you want to work with - apart from missing the mass), it is a differential equation. What is v in terms of x?

 

[Salvador_]

Of course it makes sense (the first expression is what you want to work with - apart from missing the mass), it is a differential equation. What is v in terms of x?

Sorry but I just don't know how to solve it if the left side is with respect to time and the other side is position.

 

[scottdave]

Homework Statement

A particle of mass m is subject to a force F (x) = -kx. The initial position is
zero, and the initial speed is v₀. Find x(t).

Homework Equations

F = m*v*dv/dx = -kx
v = dx/dt

Net force = mass times acceleration, but you have F = (mass)*(velocity)*(dv / dx) {dv/dt is acceleration}
Why have you set it up dv/dx?
How about just this: Net force = (mass)*(dv/dt) which is (mass)*(d²x/dt²)

 

[Chestermiller]

To expand on what @scottdave said, you have the following 2nd order ordinary differential equation with constant coefficients:$$m\frac{d^2x}{dt^2}=-kx$$Do you know how to solve this equation for x as a function of t?

 

[Orodruin]

Sorry but I just don't know how to solve it if the left side is with respect to time and the other side is position.

The time derivative of velocity is the second derivative of position. This results in a linear second order differential equation with constant coefficients.

 

[Salvador_]

Net force = mass times acceleration, but you have F = (mass)*(velocity)*(dv / dx) {dv/dt is acceleration}
Why have you set it up dv/dx?
How about just this: Net force = (mass)*(dv/dt) which is (mass)*(d²x/dt²)

I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt.
The book I've been using (Morin) was introducing me to this and I haven't done second order DE's yet.

 

[Salvador_]

Net force = mass times acceleration, but you have F = (mass)*(velocity)*(dv / dx) {dv/dt is acceleration}
Why have you set it up dv/dx?
How about just this: Net force = (mass)*(dv/dt) which is (mass)*(d²x/dt²)

I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt.

To expand on what @scottdave said, you have the following 2nd order ordinary differential equation with constant coefficients:$$m\frac{d^2x}{dt^2}=-kx$$Do you know how to solve this equation for x as a function of t?

No I've been using Morin's CM book and have only been introduced to first order DE's so far.

 

[Salvador_]

The time derivative of velocity is the second derivative of position. This results in a linear second order differential equation with constant coefficients.

Oh ok, I've been trying to solve it in first order because I haven't been introduced to second order yet.

 

[scottdave]

Can you think of a function, which when you take the derivative of it twice, gives you back the negative of the function (multiplied by a constant)? Note that the behavior of the mass/spring is to oscillate back and forth.

 

[Salvador_]

Oh yes I

Can you think of a function, which when you take the derivative of it twice, gives you back the negative of the function (multiplied by a constant)? Note that the behavior of the mass/spring is to oscillate back and forth.

I yes I recognise it's cos and sin. Thanks all.

 

[haruspex]

After integrating: (a bit messy)
t = (m/k)^1/2*arcsin(x*(k/v₀²m)^1/2) + c
c=0.

You did well to get to this, and it is quite easy from there. Just turn it around to make it x as a function of t:
(t - c)(k/m)^1/2=arcsin(x*(k/v₀²m)^1/2)
sin((t - c)(k/m)^1/2)=x*(k/v₀²m)^1/2
etc.

 

[scottdave]

Since it is messy, as you said, one good thing is to check dimensions, which may reveal possible errors
From your formula, (m/k)^1/2 needs to have the dimension of time. Knowing that Force = -kx, you can find that k has dimension of mass/time². Working that through, it does indeed have dimension of time. Then argument of arcsine needs to be dimensionless.
(x*(k/v₀²m)^1/2) It indeed does turn out to be dimensionless.

 

[scottdave]

Oh yes I

I yes I recognise it's cos and sin. Thanks all.

Were you able to get the correct answer?

 

[Salvador_]

Were you able to get the correct answer?

I'm thinking it's:
x = cos(kt)/k + v₀sin(kt)/k -1/k
So when t=0 , x=0 and v=v₀

 

[Salvador_]

I'm thinking it's:
x = cos(kt)/k + v₀sin(kt)/k -1/k
So when t=0 , x=0 and v=v₀

Wait I think this is wrong because a isn't equal to -kx then.

 

[Salvador_]

x = c*cos(k^1/2t) + v_o*sin(k^.5t)/k<sup>1/2
Would I be allowed to set c=0 to make x(0) = 0 so it'll then just be:

X = V_o*</sup>sin(k^.5t)/k^1/2

 

[Orodruin]

I'm thinking it's:
x = cos(kt)/k + v₀sin(kt)/k -1/k
So when t=0 , x=0 and v=v₀

What is the 1/k term doing there?

 

[Orodruin]

x = c*cos(k^1/2t) + v_o*sin(k^.5t)/k<sup>1/2
Would I be allowed to set c=0 to make x(0) = 0 so it'll then just be:

X = V_o*</sup>sin(k^.5t)/k^1/2

A priori, the constant in front of the sine term should be arbitrary as well. It is when you adapt to the initial conditions that these constants are fixed.

 

[scottdave]

Also, the expression for x(t) should wind up having dimension of length. This means that c*cos() should have dimension of length {cosine and sine are ratios - remember your triangles, so are dimensionless}. Also, v₀/k^1/2 needs to have dimension length. You need to know the dimensions of k. Going back to the original definition of the spring: Force = -k*x, so k has dimensions of Force / Length, which reduces to Mass / Time².
And another thing, the argument of sine( ) and cosine( ) need to be dimensionless. Your expression will not meet these requirements.

Last thing, when you think you have the answer for x(t), you can take the second derivative, then plug back into your original differential equation, to see if it satisfies it.