Underdamped motion (Formula deriving)

  • Thread starter Thread starter noppawit
  • Start date Start date
  • Tags Tags
    deriving Motion
Click For Summary
SUMMARY

The discussion focuses on deriving the equation of motion for underdamped motion, specifically the transition from the complex exponential form to the sinusoidal form. The equation of motion is established from the second-order differential equation m\ddot{x} + c\dot{x} + kx = 0, leading to the characteristic roots λ = -ζωₙ ± i√(1-ζ²). The transformation from the complex exponential solution xₕ(t) = e^{-ζωₙ t}(A sin(ωₙt + φ)) is clarified using Euler's formula, which relates complex exponentials to sine and cosine functions.

PREREQUISITES
  • Understanding of differential equations, particularly second-order linear equations.
  • Familiarity with the concepts of damping in oscillatory systems.
  • Knowledge of Euler's formula and its application in complex analysis.
  • Basic grasp of harmonic motion and natural frequency.
NEXT STEPS
  • Study the derivation of the characteristic equation for second-order differential equations.
  • Learn about the physical implications of damping ratios in mechanical systems.
  • Explore the application of Fourier transforms in analyzing oscillatory motion.
  • Investigate the relationship between complex numbers and trigonometric functions in signal processing.
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying dynamics, control systems, or mechanical vibrations, will benefit from this discussion.

noppawit
Messages
27
Reaction score
0

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
 
Physics news on Phys.org
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 :)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
958
Replies
4
Views
2K
Replies
2
Views
6K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K