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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Ask a homework question - tutors are online