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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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gomezzzzzzzzz... | Student | (Level 1) Honors

Posted January 24, 2013 at 3:48 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.

 

Write down an equation that describes the position of the oscillating mass as a function of time ( ie. what is x(t)? ).

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted May 30, 2013 at 4:54 PM (Answer #1)

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We first have to solve for the phase constant and the amplitude in order to get the position at t = 0.4s.

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

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

where:

mass = 200 g

x = length/distance = 5.0 cm

v = velocity = -30 cm/s

f = frequency = 2.0Hz

t = time = 0 s

A = amplitude

`phi` = phase constant

`omega` = angular frequency

`omega = 2*pi*f`
`omega = 2*pi*2.0`
`omega = 4pi`

 

Phase Constant

`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)` equation 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)` equation 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 from trigonometric identities: `tanx = sinx/cosx`

`6 = 4pi*tan phi`

`6/(4pi) = tan phi`

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

`phi = 25.52` -> phase constant

 

Amplitude

`A = (x)/(cos phi)`

`A = 5.0/(cos 25.52)`

`A = 5.5407`

Now we can solve the position at t = 0.4 s

 

Finally,

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

`x(t) = 5.5407 cos (4pi*0.5 + 25.52)`

x(t) = 4.8 cm

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