The angle (with respect to the vertical) of a simple pendulum is given by e = 0mcos[(5.88 rad/s)t + p]. 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 which 6 changes with thew of the SHM.) Pivot point Is= Le cose Fesine- (a) (a) Number i Unit (b) Number i Unit

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Chapter1: Units, Trigonometry. And Vectors
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The angle (with respect to the vertical) of a simple pendulum is given by e = 0mcos[(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 p and (b) the maximum angle em? (Hint: Don't confuse the rate de/dt at which e
changes with the w of the SHM.)
Pivot
point
L.
e\L
S= Le
F, cose
Fesine-
(a)
(b)
(a) Number
i
Unit
(b) Number
Unit
>
Transcribed Image Text:The angle (with respect to the vertical) of a simple pendulum is given by e = 0mcos[(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 p and (b) the maximum angle em? (Hint: Don't confuse the rate de/dt at which e changes with the w of the SHM.) Pivot point L. e\L S= Le F, cose Fesine- (a) (b) (a) Number i Unit (b) Number Unit >
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