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A force of 400 Newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. After finding the equation of motion, calculate x(t = pi/12). I know this is a second order differential equation but I don't know where to start. k=245.25 w(frequency)=2.2147 hz The general formula is x(t)=c1cos(wt) +c2sin(wt) and I think x(0)=10 but help would be really appreciated.
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I think you mean x'(0) = 10 since the initial speed is 10, not the initial position?
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