Answer:
a) x_max = 0.20794 m
b) v_max = 3.8436 m/s
c) P = 0.05883 W
Explanation:
Given:
- The stiffness k = 205 N / m
- The mass m = 0.6 kg
- initial compression of the spring xi = 13 cm
- initial speed of the mass vi = 3 m/s
Find:
(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?
Solution:
- Conservation of energy principle can be applied that the total energy U of the system remains constant. So the Total energy is:
U = K.E + P.E
U = 0.5*m*v^2 + 0.5*k*x^2
- We will take initial point with given values and maximum compression x_max when v = 0.
0.5*m*vi^2 + 0.5*k*xi^2 = 0.5*k*x_max^2
(m/k)*vi^2 + xi^2 = x_max^2
x_max = sqrt ( (m/k)*vi^2 + xi^2 ) = sqrt ( (.6/205)*3^2 + .13^2
x_max = 0.20794 m
- The angular speed w of the harmonic oscillation is given by:
w = sqrt ( k / m )
w = sqrt ( 205 / 0.6 )
w = 18.48422 rad/s
- The maximum velocity v_max is given by:
v_max = - w*x_max
v_max = - (18.48422)*(0.20794)
v_max = 3.8436 m/s
- The amount of power required to stabilize each oscillation is given by:
P = E_cycle / T
Where, E = Energy per cycle = 0.02 J
T = Time period of oscillation
T = 2π/w
P = E_cycle*w / 2π
P = (0.02*18.48422) / 2π
P = 0.05883 W
