A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an init

Question

4 years ago

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A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an init

ial compression of the spring of 13 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?

Physics

1 answer:

VikaD [51] · Answer 1 · 2022-01-15T07:20:08+03:00

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 )