The problem involves a particle of mass m subjected to a force described by F(x) = kx, where k is a positive constant. The initial conditions specify an initial position x0 and an initial speed of zero, leading to the task of finding the position function x(t) over time.
The discussion has progressed through various attempts to identify suitable functions for the differential equation. Participants have offered guidance on the nature of solutions and have explored different forms of potential solutions, including exponential functions. There is recognition of multiple independent solutions to the equation.
Participants note the original poster's limited background in differential equations, which may influence the depth of discussion and understanding of the problem-solving methods being explored.
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.Fascheue said:d^2x/dt^2 = kx/m
dx/dt = (kx^2)/(2m)
Not really, the most advanced math I’ve taken is Calculus 1, so my knowledge of differential equations is limited.TSny said: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?
OK.Fascheue said:Not really, the most advanced math I’ve taken is Calculus 1, so my knowledge of differential equations is limited.
I believe e^t should work.TSny said: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##?
Great. Can you "doctor" this solution so that it will solve ##\ddot x = \frac{k}{m} x##?Fascheue said:I believe e^t should work.
Would it be x = (m/k)e^t?TSny said:Great. Can you "doctor" this solution so that it will solve ##\ddot x = \frac{k}{m} x##?
Oh, it would just be x = (k/m)e^t wouldn’t it be?TSny said: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##?
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.Fascheue said:Oh, it would just be (k/m)e^t wouldn’t it be?
I think I see now, should it be e^sqrt(k/m)t?TSny said: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 e^-sqrt(k/m)t should work as well.TSny said:Yes. Good. It will be important to see that there is another value of ##b## that will also work.
I plugged in the constants and got x0/2 for both A and B. Is that correct?TSny said: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.
Yes. That should do it. You can express the result in terms of a hyperbolic trig function if you wish.Fascheue said:I plugged in the constants and got x0/2 for both A and B. Is that correct?