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

Question

An 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, (a) Find its position function x(t) and express your answer in the form x(t) = A cos( omega - phi ) and (b) determine the block's speed as it passes through equilibrium (note that since there is no resistance, this speed will be the same every time the objects goes 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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