Question

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

Answer 1

Answer:

(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

$T=2\pi\sqrt{\frac{m}{K}}$

$T=2\pi\sqrt{\frac{0.3}{160}}$

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

Answer 2

(a) the maximum stretch of the spring is 17.7cm.

(b) the maximum speed during the motion is 4.08 m/s.

(c) the average power input required is 0.11W.

Spring-mass system:

(a) The law of conservation of energy suggests that the total mechanical energy of the system remains conserved in case of a constant or conservative force. So,

the initial energy of the spring-mass system = final energy

also, the total energy remains conserved, so, when the potential energy is maximum, the kinetic energy is zero.

$KE_i+PE_1=KE_f+PE_f\\\\\frac{1}{2}mu^2+\frac{1}{2}kx^2=0+\frac{1}{2}kA^2$

where, u = 3 m/s is the initial speed of the mass m = 0.3kg

k = 160 N/m is the spring constant and x = 12cm = 0.12m is the initial displacement of the spring and A is the maximum displacement of the spring.

$0.3\times3^2+160\times(0.12)^2=160A^2\\\\A=0.177m=17.7cm$

(b) when the speed is maximum, the kinetic energy is maximum, so the potential energy must be zero, then:

$KE_i+PE_1=KE_f+PE_f\\\\\frac{1}{2}mu^2+\frac{1}{2}kx^2=0+\frac{1}{2}mv^2$

where v is the maximum speed.

$0.3\times3^2+160\times(0.12)^2=0.3v^2\\\\v=4.08\;m/s$

(c) the energy dissipated per cycle is 0.03 J which is equal to the work done by the system W. The average power is defined as the rate of work.

So the average power P per cycle is :

P = W/T

where T is the time period of oscillation given by:

$T=2\pi\sqrt{\frac{m}{k} } \\\\T=2\pi\sqrt{\frac{0.3}{160} }\\\\T=0.272s$

So,

$P=\frac{0.03}{0.272} W$

P = 0.11 W

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