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Unformatted text preview: Shoreline CC PHYS 122 Homework #3 Solution Chap 15, Problems 4, 10, 13, 44, 57, 76, 77, 78 4. (a) The acceleration amplitude is related to the maximum force by Newtons second law: F max = ma m . The textbook notes (in the discussion immediately after Eq. 157) that the acceleration amplitude is a m = 2 x m , where is the angular frequency ( = 2 f since there are 2 radians in one cycle). The frequency is the reciprocal of the period: f = 1/ T = 1/0.20 = 5.0 Hz, so the angular frequency is = 10 (understood to be valid to two significant figures). Therefore, = = 0.12 10 0.085 = 10 . 2 2 F m x m max kg rad / s m N b gb gb g (b) Using Eq. 1512, we obtain ( 29 ( 29 2 2 0.12kg 10 rad/s 1.2 10 N/m. k k m = = = 10. We note (from the graph) that x m = 6.00 cm. Also the value at t = 0 is x o =  2.00 cm. Then Eq. 153 leads to f = cos 1 ( 2.00/6.00) = +1.91 rad or 4.37 rad. The other root (+4.37 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0....
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This note was uploaded on 11/15/2010 for the course PHYSICS 122 taught by Professor Bertranddano during the Winter '08 term at Shoreline.
 Winter '08
 BertrandDano
 Acceleration, Force, Work

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