Question

A horizontal spring-mass system has low friction, spring stiffness 200 N/m, and mass 0.5 kg. The system is released with an initial compression of the spring of 9 cm and an initial speed of the mass of 3 m/s.

(a) What is the maximum stretch during the motion?
_ m
(b) What is the maximum speed during the motion?

  _m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
_watt

Answer 1

Given , k = 200 N/m , m = 0.5 kg , xi=initial comp, xf = maximum elongation

a) When the spring is at maximum elongation velocity is ZERO

So,

Initial energy = (1/2)*k*xi² + (1/2)*m*v²

Final energy = 0.5*k*xf²

Using conservation of energy

(1/2)*k*xi² + (1/2)*m*v²= (1/2)*k*xf²

(1/2)*200*0.09^2 + (1/2)*0.5*3^2 = (1/2)*200*xf^2

xf = 0.175 m

b) For finding maximum speed ,

Using conservaiton of energy:

(1/2)*k*xi^2  + (1/2)*m*v²= (1/2)*m*V²

(1/2)*200*0.09^2 + (1/2)*0.5*3^2 = (1/2)*0.5*V^2

V = 3.5 m/s

c) Tme period T= 2*pi*sqrt(m/k) = 2*pi*sqrt(0.5/200) = 0.314s

Now ,

Power input = 0.02/0.314= 0.0637 W