The angle (with respect to the vertical) of a simple pendulum is given by 0 = 0mCos[(5.88 rad/s)t + p]. If at t = 0, 0 = 0.0450 rad and d0/dt = -O.160 rad/s, what are (a) the phase constant p and (b) the maximum angle 0m? (Hint: Don't confuse the rate d0/dt at which 0 changes with the w of the SHM.) Pivot point L. e\L s= L0 cose Fesine (a) (b) (a) Number i Unit (b) Number i Unit

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter12: Oscillatory Motion
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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 + p]. 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 de/dt at which 0
changes with the w of the SHM.)
%3D
Pivot
point
S= Le
F,cose
Fgsine-
(a)
(b)
(a) Number
i
Unit
(b) Number
i
Unit
>
Transcribed Image Text:The angle (with respect to the vertical) of a simple pendulum is given by 0 = 0mcos[(5.88 rad/s)t + p]. 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 de/dt at which 0 changes with the w of the SHM.) %3D Pivot point S= Le F,cose Fgsine- (a) (b) (a) Number i Unit (b) Number i Unit >
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