Answer
See the detailed answer below.
Work Step by Step
a) We know that the frequency is given by
$$f=\dfrac{1}{T}\tag 1$$
where $T$ is the period and is given by
$$T=2\pi\sqrt{\dfrac{m}{k}}$$
Plugging into (1);
$$f=\dfrac{1}{2\pi\sqrt{\dfrac{m}{k}}} $$
Plugging the known;
$$f=\dfrac{1}{2\pi\sqrt{\dfrac{0.20}{10}}} =\color{red}{\bf 1.125}\;\rm Hz$$
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b) Since the spring and the block are as one unit and are oscillating without friction, the energy is conserved.
$$E_i=E_f$$
$$K_i+U_{si}=K_f+U_{sf}$$
$$ \color{red}{\bf\not} \frac{1}{2}mv_i^2+ \color{red}{\bf\not} \frac{1}{2}kx_i^2= \color{red}{\bf\not} \frac{1}{2}mv_f^2+ \color{red}{\bf\not} \frac{1}{2}kx_f^2$$
We need to solve for $x_f$ which is the distance from equilibrium when the block's speed $v_f=50$ cm/s.
$$ mv_i^2-mv_f^2+ kx_i^2= kx_f^2$$
Hence,
$$x_f=\sqrt{\dfrac{ mv_i^2-mv_f^2+ kx_i^2}{k}}$$
$$x_f=\sqrt{\dfrac{ m[v_i^2- v_f^2]}{k}+ x_i^2}$$
Plugging the known;
$$x_f=\sqrt{\dfrac{ (0.20)[1.0^2- 0.50^2]}{10}+ (-0.20)^2}=\bf0.2345\;\rm m$$
$$x_f =\color{red}{\bf 23.45}\;\rm cm$$
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c) In oscillation systems, we know that the displacement is given by
$$x_{(t)}=A\cos(2\pi ft+\phi_0)$$
At $t=0$, $x=-0.2$ m.
So
$$-0.2=A\cos(2\pi f(0)+\phi_0)$$
$$-0.2=A\cos( \phi_0)\tag 2$$
Now we need to find the amplitude and the phase constant.
We know that the amplitude is given by using the conservation of energy too.
$$ \color{red}{\bf\not} \frac{1}{2}kA^2= \color{red}{\bf\not} \frac{1}{2}mv_{\rm max}^2$$
$$ A = \left( \sqrt{ \dfrac{m}{k}}\right)v_{\rm max} \tag 3$$
where $v_{\rm max}$ is given by
$$ \color{red}{\bf\not} \frac{1}{2}mv_{\rm max}^2=\color{red}{\bf\not} \frac{1}{2}mv_i^2+ \color{red}{\bf\not} \frac{1}{2}kx_i^2$$
$$ mv_{\rm max}^2= mv_i^2+ kx_i^2$$
$$ v_{\rm max} =\sqrt{v_i^2+ \dfrac{kx_i^2}{m}}$$
Plugging into (3);
$$ A = \left( \sqrt{ \dfrac{m}{k}}\right) \sqrt{v_i^2+ \dfrac{kx_i^2}{m}}$$
Plugging the known;
$$ A = \left( \sqrt{ \dfrac{0.20}{10}}\right) \sqrt{(1)^2+ \dfrac{(10)(-0.20)^2}{0.20}}=\bf 0.2449\;\rm m$$
Plugging into (2)
$$-0.2=0.2449\cos( \phi_0) $$
Thus,
$$\phi_0=\cos^{-1}\left(\dfrac{-0.2}{0.2449}\right)=\bf \pm 2.526\;\rm rad$$
Since the cosine was negative so the object is on the third or the fourth quadrant. But since the velocity is positive when the position is negative, so in a circular motion, the object will be moving to the right from the negative side which is at the third quadrant.
Thus,
$$\phi_0=\bf-2.526\;\rm rad$$
Plugging the two results of amplitude and the phase constant into the position function above.
$$x_{(t)}=0.2449\cos(2\pi ft-2.526)$$
So, at $t=1$,
$$x_{(1)}=0.2449\cos(2\pi (1.125)(1)-2.526)=\bf -0.0414\;\rm m$$
$$x_{1}=\color{red}{\bf -4.14}\;\rm cm$$
