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 Determine the phase constant
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To solve for the phase constant, we should take note of 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
`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 equation 2 by the equation 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
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