Simple Harmonic Motion

  • Thread starter Moe*
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  • #1
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_{}b]1. Homework Statement [/b)

The position of a 39.8 g oscillating mass is given by the equation x(t)= (2.00cm)cos(5t- 4.00[tex]\pi[/tex]) where t is in seconds. Determine Velocity at t= 0.250s

M= 39.8 g
T= 1.2575 rad/s
K= 9.95e-1 N/m
initial postion= 2 cm

Homework Equations



V[tex]_{x}[/tex]= -[tex]\omega[/tex]Asin([tex]\omega[/tex]t + [tex]\phi[/tex])

The Attempt at a Solution



the attempt i just substitued
V[tex]_{x}[/tex]= -5 rad/s*2sin( 5.00(0.250)-4[tex]\pi[/tex]).... but this didn't work and i dont understand why???
 

Answers and Replies

  • #2
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for the the equations posted its supposed to be V subscript x= wAsin(wt + phi) and x(t) = (2.00cm)cos(5t - 4pi)
 
  • #3
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The position of a 39.8 g oscillating mass is given by the equation x(t)= (2.00cm)cos(5t- 4.00[tex]\pi[/tex]) where t is in seconds. Determine Velocity at t= 0.250s
What is the basic relation connecting velocity and position of any particle? Use that.
 
  • #4
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1
_{}b]1. Homework Statement [/b)

The position of a 39.8 g oscillating mass is given by the equation x(t)= (2.00cm)cos(5t- 4.00[tex]\pi[/tex]) where t is in seconds. Determine Velocity at t= 0.250s

M= 39.8 g
T= 1.2575 rad/s
K= 9.95e-1 N/m
initial postion= 2 cm

Homework Equations



V[tex]_{x}[/tex]= -[tex]\omega[/tex]Asin([tex]\omega[/tex]t + [tex]\phi[/tex])

The Attempt at a Solution



the attempt i just substitued
V[tex]_{x}[/tex]= -5 rad/s*2sin( 5.00(0.250)-4[tex]\pi[/tex]).... but this didn't work and i dont understand why???
it seems work to me...
 
  • #5
10
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Your right it does work, i guess I entered in the wrong units when i answered the question( m/s instead of cm/s).
 

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