A mass of stretches a spring . The mass is set in motion from its equlibrium position with a downward velocity of and no damping is applied. Determine the position of the mass at any time . Use as the acceleration due to gravity. Pay close attention to the units.

Complete Question:

A mass of 150 g stretches a spring 9 cm. The mass is set in motion from its equilibrium position with a downward velocity of 12 cm/s and no damping is applied. Determine the position of the mass at any time . Use as the acceleration due to gravity. Pay close attention to the units.

The position of the mass at any time = speed,

Explanation:

Mass, m = 150 g = 150/1000 = 0.15 kg

Extension, x = 9 cm = 0.09 m

According to Hooke's law, F = kx...........(1)

F = mg........(2)

Equating (1) and (2)

mg = kx

The differential equation that models the motion of a spring and a damper

Since there is no damping applied, c = 0

Solving the differential equation above by making , ,

m = ± 10.435 i

At the equilibrium position, t = 0, hence y = 0

..................(3)

A = 0

Inserting the value of A into equation (3)

...........(4)

The position of the mass at any time is equivalent to the speed of the spring

which is the first derivative of y

..........(5)

Initial speed is given as 12 cm/s = 0.12 m/s, substituting this into equation (5)

Putting the value of B in equation (5)

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