Answer
See the figure below.
Work Step by Step
We know that the position of the object that undergoes a simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)$$
where $A$ is the amplitude which is 4 cm, $\omega=2\pi f $ where $f$ is the frequency and its value is 2 Hz, $\phi_0$ is phase constant and it is on the second quadrant $\phi_0=\frac{2\pi }{3}=120^\circ$ and since $\phi_0$ is positive so the object is moving to the left.
Hence,
$$x_{(t)}=4\cos(2\pi ft+\frac{2\pi }{3})$$
$$x_{(t)}=4\cos(4\pi t+\frac{2\pi }{3})$$
We know that the period is given by $T=\dfrac{1}{f}=0.5\;\rm s$, so the initial position is at $x=-0.5 A=-0.5\times 4= 2$ cm.
We need to show 2 cycles of the motion, so $t_i=0$ s and $t_f=2T=1$ s.
See the figure below.
