Interpret the initial value problem
d 2 θ d t 2 + β 2 θ = 0 , θ ( 0 ) = θ 0 , d θ d t ( 0 ) = ω 0 {d^2 \theta \over dt^2}+\beta^2\theta=0, \ \theta(0)=\theta_0,\ {d\theta \over dt}(0)=\omega_0 d t 2 d 2 θ + β 2 θ = 0 , θ ( 0 ) = θ 0 , d t d θ ( 0 ) = ω 0 Use the notation
d 2 θ d t 2 = θ ′ ′ , d θ d t = θ ′ {d^2 \theta \over dt^2}=\theta'',\ {d \theta \over dt}=\theta' d t 2 d 2 θ = θ ′′ , d t d θ = θ ′ We get
θ ′ ′ + β 2 θ = 0 \theta''+\beta^2\theta=0 θ ′′ + β 2 θ = 0 The auxiliary equation is r 2 + β 2 = 0 r^2+\beta^2=0 r 2 + β 2 = 0 or r 2 = − β 2 , r^2=-\beta^2, r 2 = − β 2 , whose roots are ± i β . \pm i\beta. ± i β .
The general solution is
θ ( t ) = c 1 cos ( β t ) + c 2 sin ( β t ) \theta(t)=c_1 \cos(\beta t)+c_2\sin(\beta t) θ ( t ) = c 1 cos ( βt ) + c 2 sin ( βt ) Then
θ ′ = − β c 1 sin ( β t ) + β c 2 cos ( β t ) \theta'=-\beta c_1\sin(\beta t)+\beta c_2\cos(\beta t) θ ′ = − β c 1 sin ( βt ) + β c 2 cos ( βt ) The initial conditions become
θ ( 0 ) = c 1 = θ 0 \theta(0)=c_1=\theta_0 θ ( 0 ) = c 1 = θ 0
θ ′ ( 0 ) = β c 2 = ω 0 = > c 2 = ω 0 β \theta'(0)=\beta c_2=\omega_0=>c_2={\omega_0 \over \beta} θ ′ ( 0 ) = β c 2 = ω 0 => c 2 = β ω 0 Hence
θ ( t ) = θ 0 cos ( β t ) + ω 0 β sin ( β t ) \theta(t)=\theta_0 \cos(\beta t)+{\omega_0 \over \beta}\sin(\beta t) θ ( t ) = θ 0 cos ( βt ) + β ω 0 sin ( βt )
Spring system
x d i s p l a c e m e n t d x / d t v e l o c i t y m m a s s c d a m p i n g c o n s t a n t k s p r i n g c o n s t a n t F ( t ) e x t e r n a l f o r c e \begin{matrix}
x & displacement \\
dx/dt & velocity \\
m & mass \\
c & damping\ constant \\
k & spring\ constant \\
F(t) & external\ force
\end{matrix} x d x / d t m c k F ( t ) d i s pl a ce m e n t v e l oc i t y ma ss d am p in g co n s t an t s p r in g co n s t an t e x t er na l f orce If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals mass times acceleration), we have
m d 2 x d t 2 = − k x m{d^2 x \over dt^2}=-kx m d t 2 d 2 x = − k x Or
d 2 x d t 2 + k m x = 0 {d^2 x \over dt^2}+{k\over m}x=0 d t 2 d 2 x + m k x = 0 Let β 2 = k m . \beta^2=\dfrac{k}{m}. β 2 = m k . Then the solution of the initial value problem
d 2 x d t 2 + β 2 x = 0 , x ( 0 ) = x 0 , d x d t ( 0 ) = v 0 {d^2 x \over dt^2}+\beta^2x=0, \ x(0)=x_0,\ {dx\over dt}(0)=v_0 d t 2 d 2 x + β 2 x = 0 , x ( 0 ) = x 0 , d t d x ( 0 ) = v 0 is
x ( t ) = x 0 cos ( β t ) + v 0 β sin ( β t ) , β = k m x(t)=x_0 \cos(\beta t)+{v_0 \over \beta}\sin(\beta t), \beta=\sqrt{{k \over m}} x ( t ) = x 0 cos ( βt ) + β v 0 sin ( βt ) , β = m k
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