In a science museum, a 140 kg brass pendulum bob swings at the end of a 12.0 m -

Question

In a science museum, a 140 kg brass pendulum bob swings at the end of a 12.0 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s.

Part A

At exactly 12:00 noon, how many oscillations will the pendulum have completed?

Express your answer as an integer.

Part B

What is its amplitude at noon?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

b = 0.010 / 140 = 7.14 x 10^-5

w0 = sqrt(g/L) = sqrt(9.8 / 12) = 0.904 rad/s

w = sqrt(w0^2 - b^2) = w0 = 0.904 rad/s

T = 2pi / w = 6.95 sec

(A) t = 4 hrs = 4x 3600s = 14400 sec

n = t / T = 2072 oscillations

(b) A = A0 e^(-bt)

A = 1.7 e^(-7.14 x 10^-5 x 14400)

A = 0.61 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.