Energy in a spring-mass system

1. Nov 2, 2008

lebprince

1. The problem statement, all variables and given/known data
A horizontal spring-mass system has low friction, spring stiffness 250 N/m, and mass 0.4 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?

(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?
watt

3. The attempt at a solution

i was able to solve a and b right but having problems with c. Thanks for looking.

2. Nov 2, 2008

Andrusko

Well, figure out how long it takes a cycle to happen, (the period) and then divide by 0.02 and this will be the joules dissipated per second.

Then this will be the amount of joules you need to keep pumping into the system to keep it going. (Remember watts is joules per second).

Does that sound reasonable?

3. Nov 2, 2008

lebprince

Thanks for your reply. So do i find W which is equal to sqrt(K/m) and from there find the Period and do the rest? Thanks

4. Nov 2, 2008

Andrusko

Yeah. And remember:

$$\omega = 2\pi f$$

$$T = \frac{1}{f}$$

That should be enough to get to the answer.

5. Nov 2, 2008

Andrusko

Sorry I said before to divide the period by 0.2, I meant divide 0.2 by the period. If you look at it in terms of units:

$$\frac{\frac{J}{cycle}}{\frac{s}{cycle}} = \frac{J}{cycle} X \frac{cycle}{s} = \frac{J}{s}$$

Remembering when dividing fractions you just multiply by the inverse of the bottom fraction...

6. Nov 2, 2008

lebprince

I calculated W = sqrt(250/0.4) and i got 25 for w. Then w = 2*pi*f and got f = w/2*pi = 3.978 then 1/f to get T so T = 0.251 and i divided that by 0.02 and got 12.56 but the answer was incorrect.

7. Nov 2, 2008

lebprince

Thank you got it

8. Nov 2, 2008

Andrusko

Sorry I misled you... see previous post.