105. (a) From the graph, it is clear that xm= 0.30 m. (b) With F= –kx, we see kis the (negative) slope of the graph — which is 75/0.30 = 250 N/m. Plugging this into Eq. 15-13 yields =2028Tmkπ=..s(c) As discussed in §15-2, the maximum acceleration is 2221.5 10 m/s .mmmkaxxmω===×Alternatively, we could arrive at this result using am= (2π/T)2xm.
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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.