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Classical Physics
Exploring the Origin of x=e^(rt) in Simple Harmonic Motion
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[QUOTE="Freaky Fred, post: 6061512, member: 651192"] 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. d[SUP]2[/SUP]x/dt[SUP]2[/SUP] + kx/m = 0 (Homogeneous) It was considered that the solution is of the type: x = e[SUP]rt [/SUP] This is a strategy to solve this kind of second order linear equation. Then the expression is:(Just by putting e[SUP]rt[/SUP] in place of x.) d[SUP]2[/SUP](e[SUP]rt[/SUP])/dt[SUP]2[/SUP] + e[SUP]rt[/SUP]k/m = 0 Deriving: x = e[SUP]rt[/SUP] ; dx/dt = re[SUP]rt [/SUP] ; d[SUP]2[/SUP]x/dt[SUP]2[/SUP] = r[SUP]2[/SUP]e[SUP]rt[/SUP] r[SUP]2[/SUP] . e[SUP]rt[/SUP] + e[SUP]rt[/SUP]k/m = 0 We have e[SUP]rt[/SUP] in both parts. Therefore, it is one of the [B]advantages[/B] of using exponential because you will always retain this expression when deriving or integrating. e[SUP]rt [/SUP]. (r[SUP]2[/SUP] + k/m) = 0 Another [B]advantage[/B]: e^[SUP]rt[/SUP]≠0. This means that our expression r[SUP]2[/SUP] + 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[SUP]rt[/SUP]; r = ± i α) x1 = e+i αt x2 = e[SUP]- i αt[/SUP] Using Euler's formula: Another [B]advantage[/B] 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. [B][SIZE=5]Another situation[/SIZE][/B] 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 . d[SUP]2[/SUP]x/dt[SUP]2[/SUP] = -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 [SUP]-k/m . T[/SUP] vf = vi e[SUP]-k/m . T[/SUP] 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. [/QUOTE]
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Exploring the Origin of x=e^(rt) in Simple Harmonic Motion
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