Originally Posted by

**Paymemoney** Hi

The following question i am having trouble completing

The equation of motion of a mass on the end of a vertical spring is

$\displaystyle \frac{d^2y}{dt^2} +4y = 0$

(where y is the displacement of the mass from the static equilibirum position). Given that initially ( i.e at time t=0) the displacement (y) is 3 and the velocity (y') is 0, find the solution to the equation in the form $\displaystyle y=Ccos(3t-\theta)$.

This is what i have done:

$\displaystyle r^2+4 = 0$

$\displaystyle r=\sqrt{4}j$

$\displaystyle y=Acos(2t)+Bsin(2t)$

$\displaystyle y(0)=3$

$\displaystyle 3 = Acos(0)+Bsin(0)$

$\displaystyle A=3$

$\displaystyle y'=-2Asin(2t)+3Bcos(2t)$

$\displaystyle y'(0)=0$

$\displaystyle 0=-2Asin(0)+2Bcos(0)$

$\displaystyle B=0$

$\displaystyle y=3cos(2t)$

what should i do next in order to get it in the form of $\displaystyle y=Ccos(3t-\theta)$.

P.S