# B with l 185 m and x 053 m we obtain t 21 s from

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Unformatted text preview: art (a). _______________________________________________________________________ _ Problem 15-53 ((My solution)) θ 0 = 0.040rad ω0 = −0.200rad / s ω = 4.44rad / s θ = θ m cos(ωt + φ ) dθ = ( −ω )θ m sin(ωt + φ ) dt At t = 0, θ 0 = θ m cos φ ω0 = −ωθ m sin φ (Initial condition) (a) ω0 0.20 = ωθ 0 0.040 × 4.44 φ = 0.845rad tan φ = − (b) θm = θ0 0.040 = = 0.0602rad cos φ 0.664 ((WileyPlus)) 53. Replacing x and v in Eq. 15-3 and Eq. 15-6 with θ and dθ/dt, respectively, we identify 4.44 rad/s as the angular frequency ω. Then we evaluate the expressions at t = 0 and divide the second by the first: = − ω tanφ . (a) The value of θ at t = 0 is 0.0400 rad, and the value of dθ/dt then is –0.200 rad/s, so we are able to solve for the phase constant: φ = tan−1[0.200/(0.0400 x 4.44)] = 0.845 rad. (b) Once φ is determined we can plug back in to θo = θmcosφ to solve for the angular amplitude. We find θm = 0.0602 rad. _______________________________________________________________________ _ Problem 18-34 ((My solution)) m = 0....
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## This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton University.

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