GENERAL PHYSICS (PHYS 2001/20 Assignment Gradebook ORION Downloadable eTextbook

Question

GENERAL PHYSICS (PHYS 2001/20 Assignment Gradebook ORION Downloadable eTextbook ent FULL SCREEN PRINTER VERSION BACK NEXT Chapter 10, Problem 75 A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.59 s to complete one cycle. The height of each bounce above the equilbrium position is 49.7 cm. Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person? At the equilibrium position At the highest point (a) Number (b) Number (c) Number Units Units Units Version 4.24.2.4 & Sons.Inc. All Rights Reserved. A Division of John Wiley & Sons, Ins

Explanation / Answer

a) The amplitude is the maximum displacement from the equilibrium position so in this case, amplitude is given to be 49.7 cm. You probably need to convert it to meters however. So it will be 0.497 m.

b) angular frequency is represented by the character .

T = 2/

rearranging the equation gives you: = 2/T

plug in your value given for period...

= 2/1.59 = 3.95 rads/s

c) vmax = A

plug in your given amplitude and your value for the angular frequency that we obtained in part b:

vmax = 3.95 x 0.497 = 1.96 m/s

Please rate my answer if you find it helpful, good luck...

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