The angle (with respect to the vertical) of a simple pendulum is given by 0 = 0mcos[(4.64 rad/s)t + p]. If at t = 0, 0 = 0.0350 rad and de/dt = -O.200 rad/s, what are (a) the phase constant p 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 e\L S= Le F.cose Fesine- (a) (b)

Principles of Physics: A Calculus-Based Text
5th Edition
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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[(4.64 rad/s)t + ]. If at t = 0,
0 = 0.0350 rad and de/dt = -0.200 rad/s, what are (a) the phase constant p and (b) the maximum angle 0m? (Hint:
Don't confuse the rate de/dt at which 0 changes with the w of the SHM.)
Pivot
point
Is= L0
F.cose
Fgsine-
(a)
(b)
(a) Number
i
50.86
Unit
degrees
(b) Number
i
0.0554
Unit
rad
>
Transcribed Image Text:The angle (with respect to the vertical) of a simple pendulum is given by 0 = 0mcos[(4.64 rad/s)t + ]. If at t = 0, 0 = 0.0350 rad and de/dt = -0.200 rad/s, what are (a) the phase constant p and (b) the maximum angle 0m? (Hint: Don't confuse the rate de/dt at which 0 changes with the w of the SHM.) Pivot point Is= L0 F.cose Fgsine- (a) (b) (a) Number i 50.86 Unit degrees (b) Number i 0.0554 Unit rad >
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