Ch15-p065 - 1 2 m v 2 = 1 2 m ω 2 x m 2 = 1.2 J(b Using Eq 15-5 we have f = ω/2 π = 50 oscillations per second Of course Eq 15-2 can also be

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65. (a) From the graph, we find x m = 7.0 cm = 0.070 m, and T = 40 ms = 0.040 s. Thus, the angular frequency is ω = 2 π / T = 157 rad/s. Using m = 0.020 kg, the maximum kinetic energy is then
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Unformatted text preview: 1 2 m v 2 = 1 2 m ω 2 x m 2 = 1.2 J. (b) Using Eq. 15-5, we have f = ω /2 π = 50 oscillations per second. Of course, Eq. 15-2 can also be used for this....
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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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