Answer
See the detailed answer below.
Work Step by Step
From the given graph, we can see that:
$\bullet\;\; T=4\;\rm s$
$ \bullet\;\; A=10\;\rm cm$
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a)
We know that the position is given by
$$x_{(t)}=A \cos\left(\frac{2\pi}{T}t+\phi_0\right)$$
So, at $t=0$, $x_0=-5$ cm
$$-5=10 \cos\left(\frac{2\pi}{T}(0)+\phi_0\right)$$
$$-5=10 \cos\left( \phi_0\right)$$
Hence,
$$\phi_0=\cos^{-1}\left(\dfrac{-5}{10}\right)$$
$$\phi_0=\dfrac{2\pi }{3}\;\rm rad, or\;\;\dfrac{-2\pi }{3}$$
This is in the second or the third quadrant but since the particle is moving to the left it means that it is in the upper part of the circle because, in the lower half of the circle, it will be moving to the right.
Thus, the right answer here is
$$\phi_0=\color{red}{\bf\dfrac{2\pi }{3}}\;\rm rad $$
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b) We know that the velocity of our system is given by
$$v_{(t)}=-A\omega \sin\left(\frac{2\pi}{T}t+\phi_0\right)$$
And at $t=0$,
$$v_{(0)}=-\dfrac{2\pi A}{T} \sin\left(\frac{2\pi}{T}(0)+\phi_0\right)$$
$$v_{(0)}=-\dfrac{2\pi A}{T} \sin\left( \phi_0\right)$$
Plugging the known;
$$v_{(0)}=-\dfrac{2\pi (10)}{4} \sin\left( \dfrac{2\pi }{3}\right)$$
$$v_{(0)}=\color{red}{\bf-13.6}\;\rm cm/s$$
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c) The maximum speed is given by
$$v_{\rm max}=A\omega= A\left[\dfrac{2\pi }{T}\right]$$
$$v_{\rm max}= 10\left[\dfrac{2\pi }{4}\right]$$
$$v_{\rm max}=\color{red}{\bf15.7}\;\rm cm/s$$
