Questopn about phase shift SHM

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Homework Help Overview

The discussion revolves around the phase shift in simple harmonic motion (SHM), specifically focusing on the displacement equation x(t) = x_m cos(ωt + φ) and determining the phase constant φ given certain initial conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of initial conditions on the phase constant, questioning how the initial displacement and velocity influence the value of φ. There is a focus on the relationship between the cosine and sine functions in this context.

Discussion Status

Some participants have offered insights into the relationship between initial conditions and the phase constant, while others are verifying their reasoning and considering graphical representations to clarify their understanding. Multiple interpretations of the phase constant are being explored.

Contextual Notes

Participants are discussing the implications of the initial velocity being negative and how that affects the phase constant. There is an acknowledgment of potential confusion regarding the signs of the trigonometric functions involved.

Saladsamurai
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Question about phase shift SHM

The displacement of an object is given by
x(t)=x_mcos(\omega t+\phi)
If the inital displacement is 0 and the initial v is in the negative x direction, then the phase constant must be ___rads

I know that if x=o then the cosine of the phase must be 0
\cos(\omega t+\phi)=0

so\omega t+\phi=\frac{\pi}{2}

and I am stuck from here...hints?
 
Last edited:
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Initial means t=0. So x(0)=xm*cos(phi). phi=pi/2 works for that. Is v(0) negative? If so then you have a solution.
 
Dick said:
Initial means t=0. So x(0)=xm*cos(phi). phi=pi/2 works for that. Is v(0) negative? If so then you have a solution.
v is in -x direction...that is where I am most confused, does that just mean that phi has to be negative?
 
It means v=dx/dt is negative at t=0.
 
Dick said:
It means v=dx/dt is negative at t=0.
Okay...
so if v=-\omega x_m\sin(\omega t+\phi)=-
and x_m=+
t=0
then -\omega x_m\sin(\phi) is negative
so sin(phi) is positive
Thus, phi=+pi/2

Does that check out?
 
Looks ok to me. Draw a graph of cos(x) if you don't believe me.
 
Dick said:
Looks ok to me. Draw a graph of cos(x) if you don't believe me.
I will...cause I don't.
























No I won't...cause I do.
but I probably should so I can see the relationships...

Thanks again Dick.
 

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