Oscillations of a Pendulum: How Many Complete by Noon and What is the Amplitude?

AI Thread Summary
The discussion revolves around calculating the number of oscillations completed by a brass pendulum with a 15m wire by noon, starting from a 1.5m displacement at 8:00 a.m. The damping constant is noted as 0.010 kg/s, which affects the pendulum's motion. The user initially misapplied the formula for angular frequency, leading to an incorrect calculation of oscillations. Correcting this involves using the formula ω = √(g/L) while considering the damping effect. The amplitude question remains unclear, with suggestions that it may relate to the initial displacement rather than the wire length.
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Homework Statement



In a science museum, a brass pendulum bob swings at the end of a 15m long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only .010 kg/s.

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

And what is its amplitude?

Homework Equations



w=2\pif
w=\sqrt{}k/m
T=1/f

The Attempt at a Solution



I used w=\sqrt{}k/m so I had w=\sqrt{}.01/110= .009 then I plugged that into w=2pif so .009=2pif f=.0015 and found the period by doing 1/.0015 and got 659 seconds. Then I converted the 4 hours minutes and got 14,400 and then I divided 14,400/659 and got 21.85.

For the amplitude I am not sure exactly how I would get that but would it just be the length of the wire or how high it goes or something else? Any help would be great. Thanks!
 
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Greetings! It looks like you confused k for b. Remember, in a simple pendulum,

\omega = \sqrt{\frac{g}{L}},

although you will still have to factor in the damping constant.
 
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