# Second order differential equation, physics.

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?

## 1 Answer

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?

• I'm sorry, but no, the dots won't connect. Is anything of what I did above right? – martin Jan 22 '15 at 16:33
• @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. – Ron Gordon Jan 22 '15 at 16:36
• So instead of 4.71 there should be 4.71 + (9.81 * 2)? – martin Jan 22 '15 at 16:40
• @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. – Ron Gordon Jan 22 '15 at 17:04
• Please point out what numbers you're not sure what means, I'll try to clarify. – martin Jan 22 '15 at 17:47