A 2-kg block is suspended from a spring having a stiffness of 800 N/m. If the block is given an upward velocity of 4 m/s when it is displaced downward a distance of 150 mm from its equilibrium position, determine the equation which describes the motion. Assume that positive displacement is downward. x = 0.15sin(20t) + 0.2cos(20t) x = 0.1sin(20t) + 0.15cos(20t) 02cin(20+) 0.15coc(20+)

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Chapter15: Oscillations
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Problem 60AP: A mass is placed on a frictionless, horizontal table. A spring (k=100N/m) , which can be stretched...
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A 2-kg block is suspended from a spring having a stiffness of 800 N/m. If the block is given an
upward velocity of 4 m/s when it is displaced downward a distance of 150 mm from its
equilibrium position, determine the equation which describes the motion. Assume that positive
displacement is downward.
x = 0.15sin(20t) + 0.2cos(20t)
x = 0.1sin(20t) + 0.15cos(20t)
x = 0.2sin(20t) + 0.15cos(20t)
x = 0.1sin(20t) - 0.2cos(20t)
x = -0.1sin(20t) - 0.15cos(20t)
x = -0.2sin(20t) + 0.15cos(20t)
Transcribed Image Text:A 2-kg block is suspended from a spring having a stiffness of 800 N/m. If the block is given an upward velocity of 4 m/s when it is displaced downward a distance of 150 mm from its equilibrium position, determine the equation which describes the motion. Assume that positive displacement is downward. x = 0.15sin(20t) + 0.2cos(20t) x = 0.1sin(20t) + 0.15cos(20t) x = 0.2sin(20t) + 0.15cos(20t) x = 0.1sin(20t) - 0.2cos(20t) x = -0.1sin(20t) - 0.15cos(20t) x = -0.2sin(20t) + 0.15cos(20t)
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