A block of mass 6 kg is attached to a spring with spring constant 24 Nt/m. It is

Question

A block of mass 6 kg is attached to a spring with spring constant 24 Nt/m. It is pulled 6 cm below equilibrium in a resistance-free medium and set into motion with an initial velocity of 2 m/sec in the downward direction. Find its position function x(t) and express your answer in the form x(t)=Acos(wt-a) and determine the block's speed as it psses through equilibrium.

Explanation / Answer

w=(24/6)^0.5=2 x(t)=A*cos(2*t-a) differentiating, v(t)=-A*2*sin(2*t-a) at t=0,x=-0.06 =>-0.06=A*cos(a) at t=0,v=-2 =>-2=A*2*sin(a) solving, a=1.51086 A=-1.00166 therefore, x(t)=-1.00166*cos(2*t-1.51086) at equilibrium position, x(t)=0 =>t=1.54083 s therefore, v(1.54083)=1.00166*2*sin(2*1.54083-1.51086)=2.00332 m/s

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