Questopn about phase shift SHM

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SUMMARY

The discussion centers on determining the phase constant (\(\phi\)) in simple harmonic motion (SHM) when the initial displacement is zero and the initial velocity is directed negatively along the x-axis. The equation provided is \(x(t) = x_m \cos(\omega t + \phi)\). It is established that for \(x(0) = 0\), \(\phi\) must equal \(\frac{\pi}{2}\) to satisfy the cosine condition. Additionally, since the initial velocity \(v(0)\) is negative, it confirms that \(\sin(\phi)\) must be positive, reinforcing that \(\phi\) is indeed \(\frac{\pi}{2}\).

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of phase constants in oscillatory systems
  • Basic calculus, particularly differentiation for velocity calculations
NEXT STEPS
  • Study the derivation of phase constants in SHM scenarios
  • Learn about the graphical representation of cosine and sine functions
  • Explore the implications of initial conditions on oscillatory motion
  • Investigate the relationship between displacement, velocity, and phase in SHM
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Students and educators in physics, particularly those focusing on mechanics and oscillatory motion, as well as anyone seeking to deepen their understanding of phase shifts in simple harmonic motion.

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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