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 *v x *= -30 cm/s.

The phase constant

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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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