123 adding and subtracting these equations gives m1

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Unformatted text preview: 99) dx/v gives θ0 T =4 mv dv 0 dθ =4 v θ0 0 dθ 2g (cos θ − cos θ0 ) = 8 g θ0 0 √ dθ . cos θ − cos θ0 (100) (b) Using cos φ = 1 − 2 sin2 (φ/2), and making the substitution sin x ≡ sin(θ/2) sin(θ0 /2) =⇒ cos x dx = (1/2) cos(θ/2) dθ , sin(θ0 /2) (101) Figure 4 (122) where A = A/2 and B = B/2. The normal modes are obtained by setting either A or B equal to zero. Snapshots of the spring in each of these modes are shown in Fig. 6 and Fig. 7. The first mode should reproduce the result of Exercise 4.16. And indeed, since √ angle there was defined to be 2θ, the result of this exercise gives a the √ frequency of 1 − cos 2θ ω = 2ω sin θ, in agreement with Exercise 4.16. 4.34. Coupled and damped The F = ma equations are ure 6 mx 1 ¨ = mx 2 ¨ = −kx1 − k(x1 − x2 ) − bx1 , ˙ −kx2 − k(x2 − x1 ) − bx2 . ˙ (123) Adding and subtracting these equations gives m(¨1 + x2 ) + b(x1 + x2 ) + k(x1 + x2 ) x ¨ ˙ ˙ = 0, m(¨1 − x2 ) + b(x1 − x2 ) + 3k(x1 − x2 ) x ¨ ˙ ˙ = 0. (124) These are uncoupled equations for the quantities x1 + x2 and x1 − x2 . Assuming underdamping, th...
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