Question #105990
Interpret the initial value problem
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for any physical situation and hence solve the problem
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Expert's answer
2020-03-23T12:40:54-0400

Interpret the initial value problem


d2θdt2+β2θ=0, θ(0)=θ0, dθdt(0)=ω0{d^2 \theta \over dt^2}+\beta^2\theta=0, \ \theta(0)=\theta_0,\ {d\theta \over dt}(0)=\omega_0

Use the notation


d2θdt2=θ, dθdt=θ{d^2 \theta \over dt^2}=\theta'',\ {d \theta \over dt}=\theta'

We get


θ+β2θ=0\theta''+\beta^2\theta=0

The auxiliary equation is r2+β2=0r^2+\beta^2=0 or r2=β2,r^2=-\beta^2, whose roots are ±iβ.\pm i\beta.

The general solution is


θ(t)=c1cos(βt)+c2sin(βt)\theta(t)=c_1 \cos(\beta t)+c_2\sin(\beta t)

Then


θ=βc1sin(βt)+βc2cos(βt)\theta'=-\beta c_1\sin(\beta t)+\beta c_2\cos(\beta t)

The initial conditions become


θ(0)=c1=θ0\theta(0)=c_1=\theta_0

θ(0)=βc2=ω0=>c2=ω0β\theta'(0)=\beta c_2=\omega_0=>c_2={\omega_0 \over \beta}

Hence


θ(t)=θ0cos(βt)+ω0βsin(βt)\theta(t)=\theta_0 \cos(\beta t)+{\omega_0 \over \beta}\sin(\beta t)


Spring system


xdisplacementdx/dtvelocitymmasscdamping constantkspring constantF(t)external force\begin{matrix} x & displacement \\ dx/dt & velocity \\ m & mass \\ c & damping\ constant \\ k & spring\ constant \\ F(t) & external\ force \end{matrix}

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


md2xdt2=kxm{d^2 x \over dt^2}=-kx

Or


d2xdt2+kmx=0{d^2 x \over dt^2}+{k\over m}x=0

Let β2=km.\beta^2=\dfrac{k}{m}. Then the solution of the initial value problem


d2xdt2+β2x=0, x(0)=x0, dxdt(0)=v0{d^2 x \over dt^2}+\beta^2x=0, \ x(0)=x_0,\ {dx\over dt}(0)=v_0

is


x(t)=x0cos(βt)+v0βsin(βt),β=kmx(t)=x_0 \cos(\beta t)+{v_0 \over \beta}\sin(\beta t), \beta=\sqrt{{k \over m}}


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