Question

A mass of 50 g stretches a spring 3.828125 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determine the position u of the mass at any time t.

Enclose arguments of functions in parentheses. For example, sin(2x).

Assume g=9.8 ms2. Enter an exact answer.

Answer 1

mass is 50 gram = 0.05 kg

  

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stretches a spring 3.828125 cm = 0.03828125 m

so x=0.03828125

from the Hooke's law, spring constant k is

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there is no damping so damping constant is

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DE is given by

find roots

for complex roots general solution is

......................(1)

here mass is initially released from the equilibrium position.

so y(0)=0

...............put it back in equation 1

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....................(2)

take derivative

initial downward velocity of 10 cm/s,

...................put it back in equation 2

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