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Why was my initial answer wrong? I don't get what it's asking, about the correct phase angle. I tried to use the degree mode, to get the reflex angle, and convert the reflex angle to radians, but I did not get the correct answer.
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3.72 rad is correct, but I have no idea how that was solved.
Ooops I forgot to attach the question for that part. Here it is: What is the value of the phase angle ϕ if the initial velocity is positive and the initial displacement is negative?

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hint: the displacement x(t) and speed x˙(t) of the object, can be modeled by these functions: \[\Large \begin{gathered} x\left( t \right) = A\sin \left( {\omega t + \phi } \right) \hfill \\ \dot x\left( t \right) = A\omega \cos \left( {\omega t + \phi } \right) \hfill \\ \end{gathered} \] please note that the second equation is obtained from the first one, making the first derivative with respect to time of such first equation. Now from your data, we ca write: \[\Large \begin{gathered} U = \frac{1}{2}kx_0^2\quad \Rightarrow {x_0} = - \sqrt {\frac{{2U}}{k}} \hfill \\ \hfill \\ T = \frac{1}{2}m\dot x_0^2\quad \Rightarrow {{\dot x}_0} = \sqrt {\frac{{2T}}{m}} \hfill \\ \end{gathered} \] where \( \Large U,T\) are the potential and kinetic energy respectively, whereas \(\Large {x_0}, {{\dot x}_0}\) are the coordinate of position and the speed of the object at \( \Large t=0\). Now, at \( \Large t=0\), substituting, we have: \[\Large \begin{gathered} x\left( 0 \right) = A\sin \left( \phi \right) \hfill \\ \dot x\left( 0 \right) = A\omega \cos \left( \phi \right) \hfill \\ \end{gathered} \] so, after a simple substitution, we get the subsequent algebraic system: \[\Large \left\{ \begin{gathered} - \sqrt {\frac{{2U}}{k}} = A\sin \left( \phi \right) \hfill \\ \hfill \\ \sqrt {\frac{{2T}}{m}} = A\omega \cos \left( \phi \right) \hfill \\ \end{gathered} \right.\] please solve that system with respect to \( \Large \phi \), and \( \Large A \), and you will find your correct answers. Of course \( \Large A \) is the amplitude of the motion of the object

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