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

Question

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

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Hưng Khoa 1 year 2021-09-03T21:00:43+00:00 1 Answers 1 views 0

Answers ( )

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    2021-09-03T21:02:28+00:00

    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 )

                              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

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