When a 100 cm long thing aluminium rod is lodged into the ground, so that it is vertical, it requires 25.0 N of force to deflect it 20.0 cm from the rest position. The rod is then lodged into the side of a wall, and a 100 g of mass is placed at its end and it is then pulled down a distance of 20.0 cm. As the rod vibrates, it loses 50% of its initial energy every second. Draw a graph of the mass's position with respect to time for the first 2 seconds it vibrates.
F = -kx
T = 2*pi*sqrt(mass/k)
The Attempt at a Solution
From the given information, i found the spring constant and the period to be 125 N/kg and 0.057*pi Hz. And i know that the amplitude is (1/2)^t since it loses 50% per second. So the equation is currently y = (1/2)^t*sin(35.35x) - 20. My question is does this take into account the vertical acceleration since the rod is sideways on a wall.
i got the 35.35 from 0.057*pi = 2pi/B, B = 35.35.