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A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t...

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vaderrrrrr | Student | (Level 1) Honors

Posted January 27, 2013 at 1:11 AM via web

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A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vx = -30 cm/s.

 

 

The phase constant

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted April 13, 2013 at 4:46 AM (Answer #1)

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To solve this problem, first we should remember the following expressions:

`x(t) = A cos(omega*t + phi)`

`v(t) = dx/dt = -omega A sin(omega*t + phi)`

where:

  • x = length/distance = 5.0 cm
  • v = velocity = -30 cm/s
  • f = frequency = 2.0Hz
  • t = time = 0 s
  • A = amplitude
  • `omega` = angular frequency
  • `phi ` = phase constant

`omega = 2*pi*f`

`omega = 2*pi*2.0`

`omega = 4pi`

 

`x(t) = A cos(omega*t + phi)`

`5.0 = A cos (4pi*0 + phi)`

`5.0 = A cos (0 + phi)`

`5.0 = A cos(phi)`                  ` -> 1`

 

`v(t) = dx/dt = -omega A sin(omega*t + phi)`

`-30 = -4pi A sin(4pi*0 + phi)`

`-30 = -4pi A sin(0 + phi)`

`30 = 4pi A sin(phi)`              `-> 2`

 

Now divide the expression 2 by the expression 1 (2/1):

`30/5.0 = (4pi A sin(phi))/(A cos(phi))`

`6 = 4pi (sin phi)/(cos phi)`

Remember that: `tanx = sinx/cosx`

`6 = 4pi*tan phi`

 

`6/(4pi) = tan phi `

`phi = tan^-1 (6/(4pi))`

`phi = 25.52` -> phase constant

 

 

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