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
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
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.
Nov 5th 2009, 01:36 PM
Well I got
1/4x''+25x = 0
and final answer
x(t) = .15sin(10t)
but it's wrong :(
Nov 5th 2009, 02:29 PM
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).