Ive been trying to solve this problem for 2 hours now..

Suppose a weight is set in motion from a position 3 centimeters below the equilibrium position and with a downward velocity of 4 centimeters per second. (please note that the vertical number line used for position is upside down. This is a conversion from physics and it means that position below equilibrium actually correspond to a positive value.) Assume that spring stiffness and mass of the weight mean that w=2 for this system.

Write the function x(t) that gives the position of the weight as a function of time t in seconds. (Your function should consist of a sine term and a cosine term)

Originally Posted by agape801
Ive been trying to solve this problem for 2 hours now..

Suppose a weight is set in motion from a position 3 centimeters below the equilibrium position and with a downward velocity of 4 centimeters per second. (please note that the vertical number line used for position is upside down. This is a conversion from physics and it means that position below equilibrium actually correspond to a positive value.) Assume that spring stiffness and mass of the weight mean that w=2 for this system.

Write the function x(t) that gives the position of the weight as a function of time t in seconds. (Your function should consist of a sine term and a cosine term)
$x(t) = A\sin(\omega t + \phi)$

at t = 0 ...

$3 = A\sin{\phi}$

$v(t) = A\omega \cos(\omega t + \phi)$

at t = 0 ...

$4 = 2A\cos{\phi}$

from the two equations ...

$\frac{3}{2} = \tan{\phi}$

$\phi = \arctan\left(\frac{3}{2}\right)$

$3 = A\sin\left[\arctan\left(\frac{3}{2}\right)\right]$

$3 = A \cdot \frac{3}{\sqrt{13}}$

$A = \sqrt{13}$

$x(t) = \sqrt{13} \sin(2t + \phi)$

$x(t) = \sqrt{13}\left[\sin(2t)\cos{\phi} + \cos(2t)\sin{\phi}\right]$

$x(t) = \sqrt{13}\left[\sin(2t) \cdot \frac{2}{\sqrt{13}} + \cos(2t) \cdot \frac{3}{\sqrt{13}}\right]$

$x(t) = 2\sin(2t) + 3\cos(2t)$

graph of position and velocity attached ...