Qn4. A damped harmonic oscillator involves a block of mass 2.0kg and a spring with a stiffness 10 N/m. The damping force is proportioanla to the velocity of the oscillator. Initially it osicillates with an amplitude of 25cm. Due to the damping, the amplitude fallsto three-fourth of this initial value after 4 complete cycles.. (a) What is the value of the damping constant? (b) How much energy has been "dissipated" during the 4 cycles? Here are my thoughts.. Initial A=0.25m after 4 cycle it reduced to = 3/4 * 0.25 = 0.1875m w=sqrt [ k/m - (b/2m)sq] = sqrt [ 10/2 - (bsq)/16 = sqrt [(80-bsq) / 16 ] i will be using x(t) = A exp (-b/2m)t sin (wt + teta) -----------> equation 1 at t= 0 for the first cycle A= 0.25m 1 cycle = 2pi, after 4 complete cycle. it will be at 8pi therefore putting the values A= 0.25, t=8pi, m=2kg and w= sqrt [(80-bsq) / 16 ] into equation 1 0.1875= 0.25 exp (-b/4) 8pi sin (w8pi) * sqrt [(80-bsq) / 16 ] here i solve for b...to get answer the first answer am i on the right track?