Answer
See the detailed answer below.
Work Step by Step
From the given graph, we can find some data.
The period is 12 s, and the maximum speed is 60 cm/s.
And we know that the position of a particle in simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)$$
$$x_{(t)}=A\cos(2\pi f t+\phi_0)$$
where $f=1/T$, so
$$x_{(t)}=A\cos\left(\frac{2\pi}{T} t+\phi_0\right)\tag 1$$
And we know that the velocity of a particle in simple harmonic motion is given by
$$v=-A\omega\sin\left(\frac{2\pi}{T} t+\phi_0\right)$$
where $A\omega$ is the maximum speed.
So,
$$v=-v_{\rm max}\sin\left(\frac{2\pi}{T} t+\phi_0\right)\tag 2$$
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a) The amplitude is given by
$$v_{\rm max}=A\omega=A\left[\dfrac{2\pi}{T}\right]$$
So,
$$A=\dfrac{Tv_{\rm max}}{2\pi}$$
Plugging the known;
$$A=\dfrac{(12)(60) }{2\pi}=\rm 114.6\;cm\approx \color{red}{\bf 115}\;\rm cm$$
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b)
Now we need to find $\phi_0$, so at $t=0$ s, the velocity is -30 cm/s.
$$ \color{red}{\bf\not}-30= \color{red}{\bf\not}-60\sin\left(\frac{2\pi}{12} (0)+\phi_0\right) $$
$$\dfrac{1}{2}= \sin\left( \phi_0\right) $$Thus,
$$\phi_0=\sin^{-1}\left(\dfrac{1}{2} \right)=\dfrac{\pi}{6}\;\rm rad, or\;\dfrac{5\pi}{6}$$
But we can see that the initial speed of the particle is negative and it is slowing down. This means that the particle is assumed to be in the second quadrant.
So the right answer here is
$$\phi_0=\color{red}{\bf \dfrac{5\pi}{6}}\;\rm rad$$
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c) At $t=0$, its position is given by (1):
$$x_{(0)}=115\cos\left(\frac{2\pi}{12} (0)+\dfrac{5\pi}{6}\right) $$
$$x_0=\color{red}{\bf-99.6}\;\rm cm$$
