Question

A 0.810- block oscillates on the end of a springwhosespring constant is = 42.0 . The mass moves in a fluid which offers a resistive force, where = 0.610 .a) What is the period of the motion?b) What is the fractional decrease in amplitude percycle?

Answer 1

a)the angular frequency of the motion isw = (k/m)^(1/2)or (2π/T) = (k/m)^(1/2)or T = 2π * (m/k)^(1/2)m = 0.810 kg and k = 42.0 N/mb)In general,the frequency w of the driving force isvariable,whereas the natural frequency w_o of theoscillator is fixed by the values of kand m.Newton's second law inthis situation givesΣF = m * a F_osin(wt) - b *(dx/dt) - kx = m *(d²x/dt²)In this situation the solution of the above equationx = A * cos(wt + φ)where A = [(F_o/m)/((w^2 - w_o^2)^2 +(bw/m)^2)^(1/2)]and w_o= (k/m)^(1/2) is the natural frequency of theundamped oscillator (b = 0).here w = 0or A = [(F_o/m)/((0^2 - w_o^2)^2 + (b *0/m)^2)^(1/2)] = (F_o/m * w_o^2)F_o= m * gor A = (m * g/m * w_o^2) = (g/w_o^2)g = 9.8 m/s^2here A gives the fractional decrease in amplitude percycle