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A force of 400 N stretches a spring 2 meters. A mass of 50 kg is attached to the
end of the spring and is initially released from the equilibrium position with an
upward velocity of 10 m/s. Find the equation of motion
My answer is x=5sin(2t)
If youu get something different, can you show me how you got it please?
Cheers.
Last edited by mrpace (2014-08-16 16:39:17)
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hi mrpace,
OK. I get that too.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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