0
$\begingroup$

I need your input on this exercise I'm doing:

"A 2-kg mass is suspended from a string. The displacement of the spring-mass equilibrium from the spring equilibrium is measured to be 50 cm. If the mass is displaced 12 cm downward from its spring-mass equilibrium and released from rest, set up the initial value problem if no damping is present."

The spring constant is:$$k = \frac{2kg * 9.81m/s}{0.5m} = 39.24N/m$$

The additional force is $$39.24N/m * 0.12m = 4.71N$$

Which gives the differential equation: $$2y'' + 39.24y = 4.71$$

Is this correct?

$\endgroup$
0
$\begingroup$

Newton's 2nd Law: The sum of the forces equal $m a$.

Sum of the forces:

$$\underbrace{-k y}_{\text{spring force}} + \underbrace{m g}_{gravity}$$

Here, $y$ represents displacement from equilibrium. Obviously, downward is positive here. Note the minus sign on the spring force is there because the spring force opposes the direction of motion.

Initial conditions:

$$y(0) = 12 \quad \dot{y}(0) = 0$$

Can you take it from here?

$\endgroup$
  • $\begingroup$ I'm sorry, but no, the dots won't connect. Is anything of what I did above right? $\endgroup$ – martin Jan 22 '15 at 16:33
  • $\begingroup$ @fadaes: I should have pointed out two things: 1) your $y$ is with respect to the origin, not equilibrium, and 2) you forgot to include gravity as an external force. You also did not specify your initial conditions as the problem asked. $\endgroup$ – Ron Gordon Jan 22 '15 at 16:36
  • $\begingroup$ So instead of 4.71 there should be 4.71 + (9.81 * 2)? $\endgroup$ – martin Jan 22 '15 at 16:40
  • $\begingroup$ @fadaes: I think so, yes. But you would be better served using variables rather than numbers. It is really hard to know what the quantities represent when you just have a bunch of numbers floating about an equation as you do. $\endgroup$ – Ron Gordon Jan 22 '15 at 17:04
  • $\begingroup$ Please point out what numbers you're not sure what means, I'll try to clarify. $\endgroup$ – martin Jan 22 '15 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.