The angle (with respect to the vertical) of a simple pendulum is given by 0 Omcos[(5.88 rad/s)t + $]. If at t = 0,0 = 0.0450 rad and de/dt = -0.160 rad/s, what are (a) the phase constant o and (b) the maximum angle 0m? (Hint: Don't confuse the rate de/dt at whic changes with the w of the SHM.) %3D

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Oscillatory Motion
Section: Chapter Questions
Problem 29P: The angular position of a pendulum is represented by the equation = 0.032 0 cos t, where is in...
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The angle (with respect to the vertical) of a simple pendulum is given by 0
= 0mcos[(5.88 rad/s)t + ø]. If at t = 0, 0 = 0.0450 rad and d0/dt = -0.160 rad/s, what are (a) the
phase constant o and (b) the maximum angle 0m? (Hint: Don't confuse the rate d0/dt at which 0
changes with the w of the SHM.)
%3D
Pivot
point
L.
Is= Le
F.cose
Fg sine
(a)
(6)
Transcribed Image Text:The angle (with respect to the vertical) of a simple pendulum is given by 0 = 0mcos[(5.88 rad/s)t + ø]. If at t = 0, 0 = 0.0450 rad and d0/dt = -0.160 rad/s, what are (a) the phase constant o and (b) the maximum angle 0m? (Hint: Don't confuse the rate d0/dt at which 0 changes with the w of the SHM.) %3D Pivot point L. Is= Le F.cose Fg sine (a) (6)
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