# F(x) to x(t)

• Fascheue

## Homework Statement

A particle of mass m is subject to a force F(x) = kx, with k > 0. The initial position is x0, and the initial speed is zero. Find x(t).

F(x) = kx

F = ma

## The Attempt at a Solution

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F = kx

ma = kx

d^2x/dt^2 = kx/m

dx/dt = (kx^2)/(2m)

x^-2dx = (k/2m)dt

-x^-1 = kt/2m + c

-x^-1 = kt/2m + x0

x(t) = -2m/kt - 1/x0

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Welcome to PF!

d^2x/dt^2 = kx/m

dx/dt = (kx^2)/(2m)
In getting the second equation from the first, you integrated the left side with respect to t. But you integrated the right side with respect to x. This is not valid since you must do the same thing to both sides of an equation.

There are various ways you can solve the first equation. Have you studied methods of solving linear ordinary differential equations?

Fascheue
Welcome to PF!

In getting the second equation from the first, you integrated the left side with respect to t. But you integrated the right side with respect to x. This is not valid since you must do the same thing to both sides of an equation.

There are various ways you can solve the first equation. Have you studied methods of solving linear ordinary differential equations?
Not really, the most advanced math I’ve taken is Calculus 1, so my knowledge of differential equations is limited.

Not really, the most advanced math I’ve taken is Calculus 1, so my knowledge of differential equations is limited.
OK.

One approach is to guess the solution. (This is a very legitimate approach to solving differential equations!)

You need to solve ##\ddot x = \frac{k}{m} x##. So, you are looking for a function ##x(t)## such that when you differentiate it twice you get back the function multiplied by a positive constant ##\frac{k}{m}##.

Suppose for the moment we forget the constant. Can you think of a function ##x(t)## that would be a solution of ##\ddot x = x##?

Fascheue
OK.

One approach is to guess the solution. (This is a very legitimate approach to solving differential equations!)

You need to solve ##\ddot x = \frac{k}{m} x##. So, you are looking for a function ##x(t)## such that when you differentiate it twice you get back the function multiplied by a positive constant ##\frac{k}{m}##.

Suppose for the moment we forget the constant. Can you think of a function ##x(t)## that would be a solution of ##\ddot x = x##?
I believe e^t should work.

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I believe e^t should work.
Great. Can you "doctor" this solution so that it will solve ##\ddot x = \frac{k}{m} x##?

Fascheue
Great. Can you "doctor" this solution so that it will solve ##\ddot x = \frac{k}{m} x##?
Would it be x = (m/k)e^t?

That's a good try, but it doesn't actually work. If ##x(t) = \frac{m}{k}e^t##, then ##\ddot x = \frac{d^2 }{dt^2}\left( \frac{m}{k}e^t \right) = \frac{m}{k}e^t = x \neq \frac{k}{m} x##.

Can you think of another way to try to modify ##e^t##?

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Fascheue
That's a good try, but it doesn't actually work. If ##x(t) = \frac{m}{k}e^t##, then ##\ddot x = \frac{d^2 }{dt^2}\left( \frac{m}{k}e^t \right) = \frac{m}{k}e^t = x \neq \frac{k}{m} x##.

Can you think of another way to try to modify ##e^t##?
Oh, it would just be x = (k/m)e^t wouldn’t it be?

Oh, it would just be (k/m)e^t wouldn’t it be?
No. If you try it you will see that it doesn't work. Taking two time derivatives just gives back the function. It doesn't give (k/m) times the function, which would be (k/m)⋅(k/m)et = (k/m)2et.

Let's try ##x(t) = e^{bt}##, where ##b## is some constant. Can you find a value (or values) of ##b## that will work?

No. If you try it you will see that it doesn't work. Taking two time derivatives just gives back the function. It doesn't give (k/m) times the function, which would be (k/m)⋅(k/m)et = (k/m)2et.

Let's try ##x(t) = e^{bt}##, where ##b## is some constant. Can you find a value (or values) of ##b## that will work?
I think I see now, should it be e^sqrt(k/m)t?

Yes. Good. It will be important to see that there is another value of ##b## that will also work.

Fascheue
Yes. Good. It will be important to see that there is another value of ##b## that will also work.
I think e^-sqrt(k/m)t should work as well.

Right. So, you have two independent solutions to the differential equation: ##e^{\sqrt{k/m}\, t}## and ##e^{-\sqrt{k/m}\, t}## .

It is not hard to check that you can combine these to obtain a general solution in the form $$x(t) = Ae^{\sqrt{k/m}\, t} + Be^{-\sqrt{k/m}\, t}$$ where ##A## and ##B## are arbitrary constants. Hopefully, you will check that this does satisfy the differential equation.

For your problem, you can determine the values of these two constants by using the information given about the initial position and velocity. Start with the information about the initial velocity.

Fascheue
Right. So, you have two independent solutions to the differential equation: ##e^{\sqrt{k/m}\, t}## and ##e^{-\sqrt{k/m}\, t}## .

It is not hard to check that you can combine these to obtain a general solution in the form $$x(t) = Ae^{\sqrt{k/m}\, t} + Be^{-\sqrt{k/m}\, t}$$ where ##A## and ##B## are arbitrary constants. Hopefully, you will check that this does satisfy the differential equation.

For your problem, you can determine the values of these two constants by using the information given about the initial position and velocity. Start with the information about the initial velocity.
I plugged in the constants and got x0/2 for both A and B. Is that correct?

I plugged in the constants and got x0/2 for both A and B. Is that correct?
Yes. That should do it. You can express the result in terms of a hyperbolic trig function if you wish.

Fascheue