What Determines Maximum Stretch and Speed in a Spring-Mass System?

[flyboy9]

Homework Statement

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.2 kg. The system is released with an initial compression of the spring of 7 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion?
(b) What is the maximum speed during the motion?
(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?

Homework Equations

Ei = Ef + W
.5KSf^2 = .5 KSi^2 + .5mv^2

The Attempt at a Solution

(a)
.5KSf^2 = .5*165*.07^2 + .5*.2*3^2
.5KSf^2 = 1.304
s = .13m

(b)
Kf+0 = answer from part a, step 2 (in this case 1.304 J)
.5mv^2 = 1.304
v = 3.611 m/s

(c)
I was not sure what to do and have not attempted this part.

 

[anathema]

Can you please provide more information or clarify the question?

Hello,

For part (c), we are asked to calculate the average power input required to maintain a steady oscillation for the spring-mass system. This means that we need to find the average amount of energy that needs to be input into the system per unit time in order for it to continue oscillating at a constant amplitude.

To do this, we can use the fact that energy input equals energy output plus any energy dissipated. In this case, the energy input is the work done by the external force on the system, which is equal to the change in kinetic energy of the mass. We can use the equation you provided, .5mv^2 = .5KSf^2, to find the final speed of the mass after each cycle of oscillation.

Since we know that there is an energy dissipation of 0.02 J per cycle, we can subtract this from the total energy input in each cycle to find the actual change in kinetic energy. Then, we can calculate the average power input by dividing the total energy input by the total time it takes for the system to complete one cycle of oscillation.

I hope this helps. Let me know if you have any further questions.