# Springs

• Nov 5th 2009, 12:20 PM
Phyxius117
Springs
Here is the problem:

A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond its natural length by a force of 30 newtons. If the spring begins at its spring-mass equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after t seconds.

So I organized the data and find that

m = 8 kg
K = 30N/1.2M = 25 N/m
and that
External Force = 1.5 m/s

How do I approach this problem?

I think I should use the formula
mx'' + kx = F0cos(wt) but not sure how to

Thanks for the help!!
• Nov 5th 2009, 12:59 PM
jfortiv
The wording of this problem is a little tricky. The word "push" is used, and it probably makes you think "push"="force". Instead, you should think of this as an initial value problem (IVP) with an initial velocity undergoing free vibration.

So, draw your FBD and you'll get:

mx'' + kx = 0, with x(0)=0 and x'(0)=v0=1.5 m/s

Let me know if you need more help from there.

Regards (Hi)
• Nov 5th 2009, 01:36 PM
Phyxius117
Well I got

1/4x''+25x = 0

x(t) = .15sin(10t)

but it's wrong :(
• Nov 5th 2009, 02:29 PM
jfortiv
Here's a little guidance:

Assume a solution: x = A*e(lambda*t), where A and lambda are constants.

You then need to find x'' by differentiating x above twice. Plug those suckers into the equation that you started with:

mx'' + kx = 0

You'll get a product on the left hand side with one factor in lambda. That factor must be equal to zero because the other factor (A*e(lambda*t)) cannot be equal to zero (it would imply the trivial solution for x). Use it to solve for lambda (you should get two values for lambda).

Let me know where you get from that point.