452_Physics ProblemsTechnical Physics

452_Physics ProblemsTechnical Physics - 454 Oscillatory...

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

View Full Document Right Arrow Icon
454 Oscillatory Motion Substitute (3), (4) into the left side of (1) and (5) into the right side of (1); −+ + + + + =− F H G I K J +− + L N M O Q P ++ + −− kAe t b m Ae t b Ae t b Ae b m tA e t b Ae t m Ae t bt m bt m bt m bt m bt m bt m bt m 2 2 22 2 2 2 cos cos sin cos sin sin cos ωφ ωφ ω ω ωφ ωω φ ω bg Compare the coefficients of Ae t bt m + 2 cos and Ae t bt m + 2 sin : cosine-term: =− − F H G I K J −= F H G I K J =− + k b m bb m m b m m k m b m k b m 2 2 2 2 2 4 4 2 sine-term: b b =+ + = af Since the coefficients are equal, xA e t bt m 2 cos is a solution of the equation. *P15.43 The frequency if undamped would be 0 4 205 10 10 6 44 0 == × = k m . . . Nm kg s. (a) With damping ππ F H G I K J = F H G I K J F H G I K J = = 0 2 2 2 44 3 193396 002 440 2 44 0 2 700 b m f 1 s kg s 2 10.6 kg 1 s s Hz .. . . . (b) In e t bt m 0 2 cos over one cycle, a time T = 2 π , the amplitude changes from A 0 to Ae bm 0 πω for a fractional decrease of AA e A ee 00 0 3 10644 .0 00202 1 1 1 0 979 98
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