Answer
See the detailed answer below.
Work Step by Step
a) The period is given by
$$T=\dfrac{1}{f}=\dfrac{1}{2}=\color{red}{\bf 0.5}\;\rm s$$
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b) The angular frequency
$$\omega=2\pi f=2\pi (2)=\color{red}{\bf 4\pi} \;\rm rad/s$$
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c) The amplitude is given by applying the conservation of energy.
$$ \color{red}{\bf\not}\frac{1}{2}kA^2= \color{red}{\bf\not}\frac{1}{2}mv_{0x}^2+ \color{red}{\bf\not}\frac{1}{2}kx_0^2$$
$$ A^2= \dfrac{mv_{0x}^2+ kx_0^2}{k}$$
$$ A = \sqrt{\dfrac{mv_{0x}^2}{k}+ x_0^2 }$$
Recall that $k=m\omega^2$; and $\omega=2\pi f$
$$ A = \sqrt{\dfrac{ \color{red}{\bf\not}mv_{0x}^2}{ \color{red}{\bf\not}m(2\pi f)^2}+ x_0^2 }$$
$$ A = \sqrt{\dfrac{ v_{0x}^2}{(2\pi f)^2}+ x_0^2 }$$
Plugging the known;
$$ A = \sqrt{\dfrac{ (-0.30)^2}{(2\pi (2))^2}+ (0.05)^2 }=\bf 0.055407\;\rm m$$
$$A=\color{red}{\bf5.54}\;\rm cm$$
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d) Phase constant is given by from
$$x_0=A\cos\phi_0$$
So,
$$\phi_0=\cos^{-1}\left[\dfrac{x_0}{A} \right]$$
Plugging the known and recalling that this is a radian angle not in degrees, you need to be careful when using your calculator.
$$\phi_0=\cos^{-1}\left[\dfrac{5}{5.54} \right]$$
$$\phi_0=\color{red}{\bf 0.445}\;\rm rad$$
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e) The maximum speed is given by
$$v_{\rm max}=A\omega=A(2\pi f)$$
$$v_{\rm max}=5.54\times 2\pi (2)=\color{red}{\bf 69.6}\;\rm cm/s$$
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f) The maximum acceleration is given by
$$a_{x,\rm max}=A\omega^2=A(2\pi f)^2$$
$$a_{x,\rm max}= 4\pi^2 Af^2$$
Plugging the known;
$$a_{x,\rm max}= 4\pi^2 (5.54\times 10^{-2})(2)^2$$
$$a_{x,\rm max}= \color{red}{\bf 8.75}\;\rm m/s^2$$
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g) The total energy is given by
$$E_{tot}=\frac{1}{2}mv_{\rm max}^2$$
$$E_{tot}=\frac{1}{2}(0.20)(69.6\times 10^{-2})^2= \color{red}{\bf 0.048}\;\rm J$$
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h) The position at $t=0.4$ is given by
$$x_{(t)}=A \cos(2\pi ft+\phi_0)$$
$$x_{(\rm 0.4\;s)}=(5.54)\cos(2\pi (2)(0.4)+0.445)=\color{red}{\bf 3.81}\;\rm cm$$
