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Unformatted text preview: Lesson 5.1 Homework Solutions 1. Total mass M = 1280 kg, the mass of the ma m = 80, x = 0.014 m The force compressing the spring = mg = 80 9.8 = 784 N 4 1 4 784 5.6 10 0.014 5.6 10 6.61 1280 6.61 1.1 2 2 F k N m x k Hz M f Hz  = = = = = = = = = 2. m = 1.5 kg, k = 200 N.m1 , x(0) = 0.1 m and v(0) = 2.0 ms1 . (i) 200 11.6 1.5 k Hz m = = = The position of the vibrating mass can be written as x(t) = A cos( t + ) 0.1 = A cos (i) v(t) = x(t) = A sin( t + ) 2 = 11.6A sin (ii) Dividing (ii) by (i) gives: 2 tan 0.1 11.6 = , or, = 1.05 rad From (i) above, 0.1 0.2 cos( 1.05) A m = = The position function is x(t) = 0.2 cos(11.6t 1.05) (ii) 2 2 0.54 11.6 T s = = = (iii) v max =  A =  0.201 11.6 = 2.3 ms1 . Amax =  A 2 =  0.201 11.6 2 = 26.8 ms2 . (iv) 3. A = 0.5 m, = 6 rad s1 , = /6 rad (i) 3 2 3 2 1 6 1 f T z s f H = = = = = (ii) 2 1 2 , 0.5 (6 177.7 ) k k m m Nm...
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This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.
 Fall '10
 George
 Physics, Force, Mass, Work

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