s15 - P15.2 (a) (b) (c) (d) x = ( 5. cm ) cos 2t+ 00 6 v=...

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P15.2 (a) () 5.00 cm cos 2 6 xt π = + At 0 t = , 5.00 cm cos 4.33 cm 6 x ⎛⎞ == ⎜⎟ ⎝⎠ (b) 10.0 cm s si n2 6 dx vt dt + At 0 t = , 5.00 cm s v =− (c) 2 20.0 cm s cos 2 6 dv at + At 0 t = , 2 17.3 cm s a (d) 5.00 cm A = and 22 3.14 s 2 T ω === *P15.4 (a) The spring constant of this spring is 2 0.45 kg 9.8 m s 12.6 N m 0.35 m F k x = we take the x -axis pointing downward, so 0 φ = 2 12.6 kg cos 18.0 cm cos 84.4 s1 8 .0 cm cos446.6 rad 15.8 cm 0.45 kg s xA t = = (d) Now 446.6 rad 71 2 0.497 rad =×+ . In each cycle the object moves , so it has moved 418 72 cm = ( ) 71 72 cm 18 15. 8 cm 51.1 m +− = . (b) By the same steps, 2 0.44 kg 9.8 m s 12.1 N m 0.355 m k 12.1 84. 41 8 .0 cm cos443.5 rad 15.9 cm 0.44 k t m = = (e) 443.5 rad 70 2 3.62 rad =+ Distance moved ( ) 7072 cm 18 15.9 cm 50.7 m + = (c) The answers to (d) and (e) are not very different given the difference in the data about the two vibrating systems. But when we ask about details of the future, the imprecision in our knowledge about the present makes it impossible to make precise predictions. The two oscillations start out in phase but get completely out of phase. P15.6 The proposed solution sin i i v x tx t t implies velocity si nc o s ii v v t ωω + and acceleration o s s i n i i v axt v t x t t 2 x = + = (a) The acceleration being a negative constant times position means we do have SHM, and its angular frequency is . At 0 t = the equations reduce to i x x = and i so they satisfy all the requirements. vv = 439
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This note was uploaded on 08/31/2010 for the course PHYSICS 7E taught by Professor Staff during the Fall '08 term at UC Irvine.

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s15 - P15.2 (a) (b) (c) (d) x = ( 5. cm ) cos 2t+ 00 6 v=...

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