HW11_Sol_F10

A comparing the given function with eq 16 60 we

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(a) Comparing the given function with Eq. 16-60, we obtain k = π /2 and ω = 12 π rad/s. Since k = 2 π / λ , then 2 4.0m 4.0m. 2 L π π = λ = = λ (b) Since ω = 2 π f , then 2 12 rad/s, f π = π which yields 6.0Hz 24m/s. f v f = = (c) Using Eq. 16-26, we have 200 N 24 m/s /(4.0 m) v m τ μ = = which leads to m = 1.4 kg.

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Nov. 8, 2010 PHYS 2101 HW#11 WileyPlus Problem Solutions 4 (d) With 3 3(24 m/s) 9.0Hz 2 2(4.0 m) v f L = = = the period is T = 1/ f = 0.11 s.

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