Rotating Mass-Spring System

In the xyxy-coordinate system, a 1 kg\SI{1}{\kilo\gram} mass is attached to one end of a massless ideal spring. The other end of the spring is fixed at the origin. The spring constant is k=10 N/mk = \SI[per-mode=symbol]{10}{\newton\per\meter}, and the spring's unstretched length is 1 m\SI{1}{\meter}.

The mass is initially being held in the air horizontally at (x,y)=(1 m,0 m),(x,y) = (\SI{1}{\meter}, \SI{0}{\meter}), and then is released so that gravity pulls it downwards. How far is the mass from the origin right at the moment it first crosses the (vertical) yy-axis?

Details and Assumptions:

  • Give your answer in meters, to 2 decimal places.
  • There is an ambient downward gravitational acceleration of 10 m/s2\SI[per-mode=symbol]{10}{\meter\per\second\squared}.
  • For the sake of this problem, assume that the spring can be stretched to great lengths.

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