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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 )
= ( −ω )θ m sin(ωt + φ )
At t = 0, θ 0 = θ m cos φ
ω0 = −ωθ m sin φ (Initial condition) (a) ω0
ωθ 0 0.040 × 4.44
φ = 0.845rad
tan φ = − (b) θm = θ0
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.
- Spring '08