Thread: Spring differential equation?

Motion of mass hanging from a spring is modeled y''=-π^2y
Given Y(0)=5 and y(3/2)=8, find:
The particular solution of this IVP and the first time (t is positive) at which the velocity is zero.

I believe i've found the IVP and that it's y(t)=12.124sin(πt)+5cos(πt) .I'm not sure if it's correct though, but I don't know how to find the velocity. Thanks!

Your solution is not correct. Using the fact that the roots of the associated auxiliary equation are then the general solution is:



We are given:

and

Use these to determine the parameters (one of which you already have correct).

To find the velocity, compute , then equate it to zero, and solve for the first positive value of .

That's what I did for the solution. How am I not correct? I plugged in 0 for t and got that C2 was equal to 5 and then solved that when 3/2 was equal to t and 8 was equal to y(t) and got 12.124, can you help me get to the solution?

How did you get 12.124? At t = 3/2, the cosine function is 0 and the sine function is -1, so you should instead get...?

Oh, I plugged in 2/3. So, it should be -8. how would I solve for velocity given this info?

Take the general solution:



Compute the derivative to get the velocity function, equate it to zero, and use the smallest positive root.

0=-8πcost(πt)-5πsin(πt). This might sound really bad, but I dont know how to get the smallest positive root.

I recommend dividing through by then using a linear combination identity.