If you did the previous question right, you hopefully got an expression for yo. You may notice that you can simplify the differential equation a little bit: d'y k -(y – yo) dt2 т The parameter Yo now plays the roll of the "relaxed length". A better term may be "equilibrium value for y". But mathematically, it's identical to a relaxed length with the spring as the only force. We continue using this equation: y(t) = yo + A cos(wt + 4) Now, solve for A (in cm) with these parameters. Again, if you need more information, enter -100000. The parameters are: •m = 200 grams • Yo = (equilibrium value) = 40 cm • k = (spring constant) = 0.03 N/cm %3D

If you did the previous question right, you hopefully got an expression for yo. You may notice that you can simplify the differential equation a little bit: d'y k -(y – yo) dt2 т The parameter Yo now plays the roll of the "relaxed length". A better term may be "equilibrium value for y". But mathematically, it's identical to a relaxed length with the spring as the only force. We continue using this equation: y(t) = yo + A cos(wt + 4) Now, solve for A (in cm) with these parameters. Again, if you need more information, enter -100000. The parameters are: •m = 200 grams • Yo = (equilibrium value) = 40 cm • k = (spring constant) = 0.03 N/cm %3D

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter3: Oscillations

Section: Chapter Questions

Problem 3.12P

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