452_Physics ProblemsTechnical Physics

# 452_Physics ProblemsTechnical Physics - 454 Oscillatory...

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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
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