Hello i was wondering if anyone can help me with the following two parts of this problem, i cannot seem to find my notes on questions like this and im struggling, any help would be brilliant, thanks ( see the attatched picture )
for part a of the problem i'm sure it just involves resolving but then it confuses me when it is asking for the initial conditions.
Newton's second law says $F=ma$. Your force is given by the spring $F=k\Delta x$. The extent of the spring is $\Delta x=vt-x$. Therefore $$ma=m\frac{d^2x}{dt^2}=kvt-kx$$ or $$\frac{d^2x}{dt^2}+\frac{k}{m}x=\frac{kv}{m}t$$ The initial conditions are $x(0)=0$, $\frac{dx}{dt}(0)=0$, and $\frac{d^2x}{dt^2}(0)=0$. With $\omega^2=\frac{k}{m}$, the solution has the form $x(t)=A\cos{\omega t}+B\sin{\omega t}+C+Dt$. The last term comes from the right side of the equation of motion