ch15-p105 - 105. (a) From the graph, it is clear that xm =...

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105. (a) From the graph, it is clear that x m = 0.30 m. (b) With F = – kx , we see k is the (negative) slope of the graph — which is 75/0.30 = 250 N/m. Plugging this into Eq. 15-13 yields =2 028 T m k π= .. s (c) As discussed in § 15-2, the maximum acceleration is 22 2 1.5 10 m/s . mmm k axx m ω === × Alternatively, we could arrive at this result using a m = (2 π / T ) 2 x m .
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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