446_Physics ProblemsTechnical Physics

446_Physics ProblemsTechnical Physics - 448 P15.23...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
448 Oscillatory Motion P15.23 Model the oscillator as a block-spring system. From energy considerations, vxA 22 2 += ωω vA max = ω and v A = 2 so A xA 2 2 2 2 F H G I K J From this we find 3 4 = and == ± 3 2 260 . cm where A = 300 . c m P15.24 The potential energy is Uk xk A t s 1 2 1 2 2 cos a f . The rate of change of potential energy is dU dt kA t t kA t s =− = 1 2 2 1 2 2 cos sin sin af . (a) This rate of change is maximal and negative at 2 2 π t = , 2 ωπ t =+ , or in general, 2 tn for integer n . Then, n = + 4 41 4360 1 a f ej s For n = 0, this gives t = 0218 s while n = 1 gives t = 109 s . All other values of n yield times outside the specified range. (b) dU dt kA s max .. . . × = −− 1 2 1 2 3 24 5 00 10 3 60 14 6 2 1 Nm m s mW bg e j Section 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online