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posted by  Chrissy0022 on 11/5/2009 2:38:24 AM  |  status: Live  |  Earned Karma: 404

simple harmonic vertical pitching motion

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General Physics N/A N/A N/A 11/5/2009 at 1:00:00 PM
Question Details:

1.      The bow of a destroyer undergoes simple harmonic vertical pitching motion with a period of 12.0 s and an amplitude of 2.5 m

a.       A.What is the maximum vertical velocity of the destroyer’s bow?

b.      B.What is the magnitude of the maximum acceleration?

c.       C. A 75 kg sailor is standing on a scale in the bunkroom in the bow.  What are the maximum and minimum readings on the scale in N?
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posted by Mobiusdick on 11/5/2009 3:04:16 AM  |  status: Live
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Response Details:
A) V = Aω sin(ωt) (maxed when sin is 1)
     ω = 2π/T = π/6
     A = 2.5
     V = 1.31m/s

B) a = A(ω)2 sin(ωt) (maxed when sin is 1)
     a = .685 m/s

C) The maximum is when the ship accelerates down, so the total acceleration on the sailor is
   atot = 9.8 + .685 = 10.485m/s2
   F = 10.485*75 = 786.375N

The minimum is when the ship accelerates upwards,
   amin = 9.8 - .685 = 9.115
   F = 9.115*75 = 683.625N
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