A simple pendulum with mass m = 1.9 kg and length L = 2.39 m hangs from the ceil

Question

A simple pendulum with mass m = 1.9 kg and length L = 2.39 m hangs from the ceiling. It is pulled back to an small angle of = 9.9° from the vertical and released at t = 0.

1)

What is the period of oscillation?
s

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2)

What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0?
N

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3)

What is the maximum speed of the pendulum?
m/s

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4)

What is the angular displacement at t = 3.5 s? (give the answer as a negative angle if the angle is to the left of the vertical)
°

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5)

What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position?
m/s2

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6)

What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?
m/s2

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7)

Which of the following would change the frequency of oscillation of this simple pendulum?

increasing the mass

decreasing the initial angular displacement

increasing the length

hanging the pendulum in an elevator accelerating downward

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(Survey Question)

8)

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Explanation / Answer

T = 2L/g = 3.10 s ANS-1
mg(sin 9.9°) = (1.9)(9.8)(sin 9.9°) = 3.2 N ANS-2

V = 2gh {h = L - L(cos 9.9°) = 2.39 - 2.35 = 0.04}
V = 2(9.8)(0.04) = 0.784 = 0.885 m/s ANS-3

at t = 3.5 s pendulum has swung one full period plus (3.5 - 3.1) = 0.4 s
0.4 s is 0.4/3.2 = 1/8 cycle
pendulum swings 9.9° every 1/4 cycle which is every 3.2/4 = 0.8 s
pendulum swings 9.9/2 = 4.95° every 1/8 cycle which is 0.4 s
thus pendulum at t = 3.5 s is displaced 4.95° from its t = 0 (starting position) ANS-4

As the only acceleration acting on pendulum is that of gravity which acts vertically, the size of the tangential acceleration (of gravity) acting on pendulum in its VERTICAL position = 0 ANS-4

By the logic of the above statement,
the size of the radial acceleration at pendulum's VERTICAL position = g ANS-5

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