Underdamped motion (Formula deriving)

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The discussion focuses on deriving the equation of motion for underdamped motion, starting from the differential equation m\ddot{x} + c\dot{x} + kx = 0. The solution involves the characteristic equation leading to the expression for the homogeneous solution x_{h}(t) using complex exponentials. The transition from the complex exponential form to the sine and cosine representation is questioned, specifically how to derive x_{h}(t) = e^{-\zeta\omega_{n} t}(A sin(\omega_{n}t + \phi)). Participants reference Euler's formula and the need for real solutions to explain the transformation. The conversation highlights the mathematical relationships necessary for understanding underdamped motion in physics.
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Homework Statement


I'm trying to derive the equation of motion of underdamped motion.


Homework Equations


From F = m\ddot{x}
-kx - c\dot{x} = m\ddot{x}
m\ddot{x} + c\dot{x} + kx = 0

x_{h}(t) = ae^{\lambda t}
\dot{x}(t) = \lambda ae^{\lambda t}
\ddot{x}(t) = \lambda^2 ae^{\lambda t}

Therefore, for underdamped: \lambda = -\zeta\omega_{n} \pm i\sqrt{1-\zeta^{2}}
x_{h}(t) = a_{1}e^{(-\zeta\omega_{n} + i\sqrt{1-\zeta^{2}}) t} + a_{2}e^{(-\zeta\omega_{n} - i\sqrt{1-\zeta^{2}}) t}
x_{h}(t) = e^{-\zeta\omega_{n} t}(a_{1}e^{i\sqrt{1-\zeta^{2}}) t} + a_{2}e^{- i\sqrt{1-\zeta^{2}}) t}) <---- From here (1)
x_{h}(t) = e^{-\zeta\omega_{n} t}(A sin(\omega_{n}t + \phi)) <---- To here (2)

From (1) to (2), how can it become like that? I tried from Euler Equation that e^{i\theta} = cos(\theta) + i sin(\theta), but I still cannot derive from (1) to (2)

P.S.\omega_{d} is damped natural frequency.

Noppawit
 
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hi noppawit! :smile:

Peix + Qe-ix

= ((P+Q)/2)cosx + i((P-Q)/2)sinx …

since we know the solution must be real, we can assume P+Q and i(P-Q) are real,

so this is in the form Rsinx + Scosx, which we can rewrite as √(R2 + S2)sin(x + tan-1S/R) :wink:
 
tiny-tim said:
hi noppawit! :smile:

Peix + Qe-ix

= ((P+Q)/2)cosx + i((P-Q)/2)sinx …

since we know the solution must be real, we can assume P+Q and i(P-Q) are real,

so this is in the form Rsinx + Scosx, which we can rewrite as √(R2 + S2)sin(x + tan-1S/R) :wink:

Thank you :)
 
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