A force of 720 newtons stretches a spring 4 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion. x(t) =-6- sin(21) x m

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter16: Oscillations
Section: Chapter Questions
Problem 28PQ: We do not need the analogy in Equation 16.30 to write expressions for the translational displacement...
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A force of 720 newtons stretches a spring 4 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.

\[ x(t) = -6 \cdot \sin(2t) \, \text{m} \]
Transcribed Image Text:A force of 720 newtons stretches a spring 4 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion. \[ x(t) = -6 \cdot \sin(2t) \, \text{m} \]
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