1. ## Another Application Problem

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

2. 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
Nothing it can't be done, because the solution has period $\displaystyle \pi$ and the target form has period $\displaystyle 2\pi/3$

3. well according to my book's answer it can be done.

This is what the answer is: $\displaystyle y=5cos(2t-53^{\circ} 08')$ or $\displaystyle y=5cos(2t-0.9271^{c})$

4. Which is not the form the question wanted it in. Seems like a typo in the question to me.

5. Originally Posted by Paymemoney
well according to my book's answer it can be done.

This is what the answer is: $\displaystyle y=5cos(2t-53^{\circ} 08')$ or $\displaystyle y=5cos(2t-0.9271^{c})$
Asserting that it can be done and then showing it not done! Nice, I will stick with as asked it can't be done. If it had asked for the solution in the form $\displaystyle y=A\cos(\omega t+\phi)$ of course my answer would be different.

Also, don't mix degrees and radians (here you should only be using radians)

CB

6. checked the answers with lecturer and he said that the book's incorrect.