Totally lost here The equation of motion for a damped oscillator is m [(d2x)/(dt

Question

Totally lost here

The equation of motion for a damped oscillator is m [(d2x)/(dt2)] = -kx - b [(dx)/(dt)] For a critically damped oscillator, b2 = 4mk: then the general solution of the differential equation is x(t) = (C1+C2t) e -bt/2m Derive the solution x(t) with initial values x(0)=0 and v(0)=v0 (at rime t=0). Determine the maximum displacement, in terms of m, k, v0. Determine the time of maximum displacement. Suppose the initial velocity were twice as large. How would the maximum displacement and the time of maximum displacement change?

Explanation / Answer

mx'' = - kx - bx'

mx'' + bx' +kx =0

x'' + b/m x' + k/m *x = 0

characteristic equation becomes

D^2 + (b/m) D + (k/m) = 0

Now this is quadratic in D and has two roots ;

so solution is x = Ae^k1*t +Be^k2*t .....Where k1 and k2 are the roots of above quadratic equation and A and B are Constants

For the case of Critically Damped we have two equal and real roots ;

(b/m)^2 - 4*(k/m) = 0

b^2 = 4*k*m

and those roots are

(-b/m+0)/2 = -b/2m

so General solution for this case

is

(A + Bt)e^kt

where k is the root of characteristic equation here k = -b/2m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.