1.) A block with a mass of 0.241 kg is attached to a spring of spring constant 4

Question

1.) A block with a mass of 0.241 kg is attached to a spring of spring constant 417 N/m. It is sitting at equilibrium. You then pull the block down 3.60 cm from equilibrium and let go. What is the amplitude of the oscillation? 2.) A block with a mass of 0.482 kg is attached to a spring of spring constant 417 N/m. It is sitting at equilibrium. You then pull the block down 3.60 cm from equilibrium and let go. What is the amplitude of the oscillation? 3.) A block with a mass of 0.241 kg is attached to a spring of spring constant 834 N/m. It is sitting at equilibrium. You then pull the block down 3.60 cm from equilibrium and let go. What is the amplitude of the oscillation? 4.) A block with a mass of 0.241 kg is attached to a spring of spring constant 417 N/m. It is sitting at equilibrium. You then pull the block down 7.20 cm from equilibrium and let go. What is the amplitude of the oscillation?

Explanation / Answer

You seem to be on the right track. w = sqrt(k/m) = 1.323 radians/sec x = A cos(wt), since the initial condition is full displacement at t=0 thus v = -Aw sin(wt), and substituting the known conditions -1.5 m/sec = -A (1.323) sin( (1.323)(0.9) ) = -A (1.229) Therefore A = 1.22 meters For (b), the total energy is simply (1/2)kA^2

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