Why is x=e^(rt) in SHM?

[velvetmist]

Main Question or Discussion Point

This may be a fool question, but i can't figure where does this come from. I would really appreciate if someone can help me. Thanks.

 

[gleem]

The answer should have been X= e^i√(k/m)t note the i

 

[Freaky Fred]

Hello!
First, Newton's second law. (mass m)
F = ma
-kx = ma
ma + kx = 0
You can divide all by mass.
a + kx/m = 0
We have a differential equation.
d²x/dt² + kx/m = 0 (Homogeneous)

It was considered that the solution is of the type: x = e^rt
This is a strategy to solve this kind of second order linear equation.
Then the expression is:(Just by putting e^rt in place of x.)

d²(e^rt)/dt² + e^rtk/m = 0

Deriving: x = e^rt ; dx/dt = re^rt ; d²x/dt² = r²e^rt

r² . e^rt + e^rtk/m = 0

We have e^rt in both parts. Therefore, it is one of the advantages of using exponential because you will always retain this expression when deriving or integrating.

e^rt . (r² + k/m) = 0

Another advantage: e^^rt≠0. This means that our expression r² + k / m = 0

r = sqrt(-k/m)

in this case, k / m is constant. For simplicity, I will say that: sqrt (k / m) = α

r = ± i α
i = sqrt(-1) = imaginary number
So we have two expressions: (using x = e^rt; r = ± i α)

x1 = e+i αt
x2 = e^{- i αt}

Using Euler's formula: Another advantage of e

x1 = cos(αt) + i sin(αt)
x2 = cos(-αt) + i sin(-αt)

Putting in the general solution of this type of equation:

x = c1 x1 + c2 x2
c1 and c2 are constants.

x = c1 ( cos(αt) + i sin(αt) ) + c2 ( cos(-αt) + i sin(-αt) )
x = c1 ( cos(αt) + i sin(αt) ) + c2 ( cos(αt) - i sin(αt) )
x = c1 cos(αt) + c1 i sin(αt) + c2 cos(αt) - c2 i sin(αt)
x = (c1 + c2) cos(αt) + (c1 - c2) i sin(αt)

Only the real part matters here.

x(t) = (c1+c2) cos(αt + δ)
x(t) = Acos(αt + δ)

δ is used to indicate the initial phase. Example: When δ = 0,it means that at time 0 its x is the farthest from the equilibrium position.

x (t) = Acos (αt + δ)
x (0) = Acos (α.0 + 0) = A.


The e is very useful for this type of calculation. you may notice some characteristics through it. Example: When the e is accompanied by an imaginary number, wait for an oscillation.

Another situation
In a more "manual" way, you can see the math mechanics that do this e appears.
For example in the drag force. Imagine a body in horizontal motion in which only drag force acts on it.
Drag Force = -kv ; k = constant ; v = speed

F = ma
-kv = ma

m . d²x/dt² = -kv

dv/dt . 1/v = -k/m

dv . 1/v = -k/m . dt

∫1/v . dv = -∫k/m . dt
limits of integration: initial speed (vi) to final speed (vf).
limits of integration: Initial time(I will consider equal to 0) to final time(T)

ln(vf) - ln (vi) = -k/m T

ln(vf/vi) = -k/m . T

vf/vi = e ^{-k/m . T}

vf = vi e^{-k/m . T}

But this way of solving, in more complex problems, can be very laborious.
I hope I've helped. If I made a mistake, please correct me.

 

[velvetmist]

It was considered that the solution is of the type: x = ert
This is a strategy to solve this kind of second order linear equation.

I understand the rest of your argument, but this was my original question and i still not getting why this is a possible solution, i mean, i can´t made a proper demostration or something. Is like i'm not taking into account the constants that integrals implies.

 

[Nugatory]

I understand the rest of your argument, but this was my original question and i still not getting why this is a possible solution, i mean, i can´t made a proper demostration or something. Is like i'm not taking into account the constants that integrals implies.

Are you asking where $x=e^{rt}$ came from?

It's an educated guess as to the form of the solution. You look at $\frac{d^2x}{dx^2}+\frac{k}{m}x=0$, you see that this can only work if the second time derivative of $x$ is a multiple of $x$, you remember that $x=e^{rt}$ has this property so you try substituting that into the equation and see if it works. Differential equations are solved this way so often that there's even a word for the initial educated guess as to the form of the solution: "ansatz".

 

[velvetmist]

Are you asking where $x=e^{rt}$ came from?

It's an educated guess as to the form of the solution. You look at $\frac{d^2x}{dx^2}+\frac{k}{m}x=0$, you see that this can only work if the second time derivative of $x$ is a multiple of $x$, you remember that $x=e^{rt}$ has this property so you try substituting that into the equation and see if it works. Differential equations are solved this way so often that there's even a word for the initial educated guess as to the form of the solution: "ansatz".

Thank you so much! I feel pretty silly now tbqh.