A wagon with the mass $m = 0.4 \,\rm{Kg}$ can move on a horizontal air path. In this way, the friction between the carriage and the ground is practically zero. We attach a spring with spring constant $k = 1.6 \pi^ 2 \, \rm{N m^{-1}}$ between the carriage and the end of the track. At the time $t = 0$, the carriage passes the equilibrium position with the speed $0.5 \pi \,\rm{m s^{-1}}$.
Find the acceleration in the points on the path where the carriage changes direction.
I found out that $y = \frac 14 \sin(2\pi x)$, but now I am stuck. Can someone give me some hints?
You have the differential equation
$$m\ddot{x}+kx=0$$ $$\implies x(t)=c_1\sin \omega t + c_2\cos \omega t$$
In which $\omega = \sqrt{k/m}=2\text{Hz}$.
At $t=0$, we pass through the equilibrium point $x=0$ hence. $$x(t=0)=0=c_2$$
At $t=0$, we have a velocity of $0.5\pi \text{ m/s}$. The velocity is the first time derivative of the position hence
$$\dot{x}(t=0)=0.5\pi\text{ m/s}=c_1\omega \cos (\omega \cdot 0)=c_1\omega.$$
We have found $c_2=0$ and $c_1=0.5\pi/2 \text{ s}=0.25\text{ s}$.
Now, we need to find the acceleration in the points in which the chart changes its direction (or velocity). This means we have to find the time for which $\dot{x}=0$ and plug this into $\ddot{x}$ to obtain the accelerations.
Hence, we have $\dot{x}=c_1\omega \cos \omega t=0 \implies \omega t = \pi/2$. Solved for $t_1=\pi/2\omega$ or $t_2=3\pi/2\omega$.
The acceleration is given by $\ddot{x}=c_1\omega^2\sin \omega t$. So we obtain $\ddot{x}_1=c_1\omega^2\sin \pi/2 = c_1\omega^2$ and $\ddot{x}_2=c_1\omega^{2}\sin 3\pi/2=-c_1\omega^{2}$ as accelerations.