11.39: A.6.1- Section 6.1 Answers
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1. y=3cos4√6t−12√6sin4√6t ft
2. y=−14cos8√5t−14√5sin8√5t ft
3. y=−1.5cos14√10t cm
4. y=14cos8t−116sin8t ft;R=√1716 ft;ω0=8 rad/s;T=π/4 s;ϕ≈−.245 rad≈−14.04∘
5. y=10cos14t+2514sin14t cm;R=√514√809 cm;ω0=14 rad/s;T=π/7 s;ϕ≈.177 rad≈10.12∘
6. y=−14cos√70t+2√70sin√70t m;R=14√6735 m;ω0=√70 rad/s;T=2π/√70 s;ϕ≈2.38 rad≈136.28∘
7. y=23cos16t−14sin16t ft
8. y=12cos8t−38sin8t ft
9. .72 m
10. y=13sint+12cos2t+56sin2t ft
11. y=165(4sint4−sint)
12. y=−116sin8t+13cos4√2t−18√2sin4√2t
13. y=−tcos8t−16cos8t+18sin8t ft
14. T=4√2 s
15. ω=8 rad/sy=−t16(−cos8t+2sin8t)+1128sin8t ft
16. ω=4√6 rad/s;y=−t√6[83cos4√6t+4sin4√6t]+19sin4√6t ft
17. y=t2cos2t−t4sin2t+3cos2t+2sin2t m
18. y=y0cosω0t+v0ω0sinω0t;R=1ω0√(ω0y0)2+(v0)2;cosϕ=y0ω0√(ω0y0)2+(v0)2;sinϕ=v0√(ω0y0)2+(v0)2
19. The object with the longer period weighs four times as much as the other.
20. T2=√2T1, where T1 is the period of the smaller object
21. k1=9k2, where k1 is the spring constant of the system with the shorter period.