440_Physics ProblemsTechnical Physics

440_Physics ProblemsTechnical Physics - 442 Oscillatory...

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442 Oscillatory Motion Section 15.2 Mathematical Representation of Simple Harmonic Motion P15.2 (a) xt =+ F H G I K J 500 2 6 .c o s cm af π At t = 0, x = F H G I K J = 6 433 o s . cm cm (b) v dx dt t == + F H G I K J 10 0 2 6 .s i n cm s bg At t = v =− . c m s (c) a dv dt t + F H G I K J 20 0 2 6 o s cm s 2 ej At t = a 17 3 . cms 2 (d) A = . cm and T === 22 2 314 ω s P15.3 400 300 o s . m a f ππ Compare this with xA t cos ωφ to find (a) ωπ 23 0 0 f . or f = 150 H z T f 1 0667 s (b) A = m (c) φπ = rad (d) = 0250 175 283 .. c o s . . s m m a f a f *P15.4 (a) The spring constant of this spring is k F x = 045 98 035 12 6 . . kg m s m Nm 2 we take the x -axis pointing downward, so φ = 0 t cos . cos . c o s . . 18 0 12 6 84 4 18 0 446 6 15 8 cm kg 0.45 kg s s cm rad cm 2 (d) Now 446 6 71 2 0 497 rad rad =× + . In each cycle the object moves 418 72 = cm, so it has moved 71 72 18 15 8 511 cm cm m a f +− = . (b) By the same steps, k 044 0355 12 1 . . kg m s m 2 k m t = = cos . cos . . ..c o s . . 18 0 12 1 84 4 18 0 443 5 15 9 cm cm rad cm (e) 443 5 70 2 3 62 rad rad Distance moved
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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