# Exploring the Origin of x=e^(rt) in Simple Harmonic Motion

• I
• velvetmist
In summary, the conversation discusses the derivation and use of the solution x = ert for second order linear equations, specifically in the context of Newton's second law and the drag force equation. This solution, which involves the use of the exponential function, is considered to be a useful and efficient strategy for solving these types of equations. The conversation also touches on the role of constants in integrals and the process of educated guessing in solving differential equations.

#### velvetmist

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.

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

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 = ert
This is a strategy to solve this kind of second order linear equation.
Then the expression is:(Just by putting ert in place of x.)

d2(ert)/dt2 + ertk/m = 0

Deriving: x = ert ; dx/dt = rert ; d2x/dt2 = r2ert

r2 . ert + ertk/m = 0

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

ert . (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 = ert; 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.

Last edited:
Freaky Fred said:
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.

velvetmist said:
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
Nugatory said:
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.

## What is the origin of the equation x=e^(rt) in simple harmonic motion?

The equation x=e^(rt) is derived from the differential equation for simple harmonic motion, which describes the motion of an object under the influence of a restoring force that is proportional to its displacement from equilibrium.

## How is the equation x=e^(rt) related to simple harmonic motion?

The equation x=e^(rt) represents the displacement of an object undergoing simple harmonic motion at any given time t. It shows that the displacement is directly proportional to the exponential function of time, with the constant r representing the frequency of oscillation.

## Why is the exponential function used in the equation x=e^(rt)?

The exponential function is used to represent the behavior of the displacement in simple harmonic motion because it is the solution to the differential equation for this type of motion. It accurately describes the oscillatory behavior of the displacement over time.

## What does the variable r represent in the equation x=e^(rt)?

The variable r represents the angular frequency of the oscillations in simple harmonic motion. It is related to the period and frequency of the motion by the equations T=2π/r and f=1/T.

## Can the equation x=e^(rt) be applied to all types of simple harmonic motion?

Yes, the equation x=e^(rt) can be applied to all types of simple harmonic motion, as long as the motion is described by a differential equation of the form x''+r^2x=0, where x is the displacement and r is a constant. This includes pendulum motion, spring motion, and other types of oscillatory motion.