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 50 cms, 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.

u(t)=     m

When does the mass first return to its equilibrium position?

Enter an exact answer.

t=     s

Answer 1

mass is 50 grams so, m =0.05 kg

spring stretches 3.828125 cm = 0.03828125 m

so x=3.828125

from the Hooke's law, spring constant k is

.

there is no damping so damping constant is c=0

..

DE is given by

find roots

for complex roots general solution is

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

here mass is initially released from rest from the equilibrium position.

so y(0)=0

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

.

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

take derivative

initial downward velocity is 50 cms so y'(0)=0.5 m/s

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

.

or   

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

.

take u(t)=0

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