A community for students.
Here's the question you clicked on:
 0 viewing
Agent_A
 one year ago
See photos below.
Agent_A
 one year ago
See photos below.

This Question is Closed

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.0Why 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.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.03.72 rad is correct, but I have no idea how that was solved.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.0Ooops 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3hint: 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.