A body with mass $m = \frac{1}{2}$ = kilogram $\left(kg\right)$ is attached to the end of a spring that is stretched two meters by a force of $100$ Newtons. It is set in motion with an initial position of $y_0=\frac{1}{2}$$\left(m\right)$ and velocity of $v_0=\frac{1}{2}$$\left(\frac{m}{s}\right)$. Find the position function of the body as well as the amplitude, frequency, period of oscillation, and time lag of the motion.
So I organized the information that I know:
$m=\frac{1}{2}kg$
$F=100N$
$y_0=\frac{1}{2}m$
$v_0=\frac{1}{2}\frac{m}{s}$
So $\frac{F}{m}=a \Rightarrow\frac{100N}{\frac{1}{2}kg} = 200\frac{m}{s^2}$
So now I have my $F=ma$ and I assume from here I integrate my acceleration to get my velocity function and then integrate my velocity function to get my position function?
Integrating my $a(t)$ I get:
$\int\frac{200m}{s^2}dt = \frac{200m}{s}+C_1$
And then integrating this again:
$\int\frac{200m}{s}+C_1dt = 200m+C_1t+C_2$
Is this correct? If so how do I find the amplitude, frequency, period of oscillation, and time lag from here?
I'm having a hard time understanding the setup, but at least the integration seems off. You are assuming that the acceleration $a$ is not a function of time. This gives you no oscillations what so ever.
From what I remember, these problems always made use of hooks laws ($F = k\Delta x$) where $\Delta x$ is the displacement from the equilibrium. If you know the force $F = 100 N$ at displacement $\Delta x = 2m$, then you can determine the constant $k$, and get another expression for the accelleration $$F = - k \Delta x = m a \rightarrow a = a(t) = - \frac{k\Delta x(t)}{m}$$
Does this sound familiar?
If your still stuck, write the equation as: $$ a(t) = -\frac{k\Delta x(t)}{m} ~ \rightarrow ~~ x'' + \frac{k}{m}x = 0 $$ which has a periodic solution, since $k, m > 0 $.