A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an initial compression of the spring of 12 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 0.03 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation? watt

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

newton's first law of motion

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