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# Sol-Ch15 - CHAPTER 15 E xercises 1 cos 5/3 sin/6 2(a(20 t/4...

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Exercises 1. cos ( θ + 5π/3), sin( θ + π/6) 2. (a) (20πt + π/4) = π/2, thus t = 1/80 s (b) v = ω A cos( ω t + φ ). Need (20πt + π/4) = 2π, so t = 7/80 s (c) a = – ω 2 A sin( ω t + φ ), Need (20πt + π/4) = 3π/2, so t = 5/80 s 3. (a) k = ΔF / Δx = 1.47 × 10 5 M/m; (b) T = 2 π(m/k) 1/2 = 0.656 s. 4. (a) a = – ω 2 x = –11.5 m/s 2 (b) sin(12t + 0.2) = 0.5, so (12t + 0.2) = π/6, 5π/6 But v < 0 means cos(12t + 0.2) < 0, so use 5π/6 which gives t = 0.201 s 5. v = –1.26 sin(3.6t – 0.5) = ±0.5(1.26) thus (3.6 t – 0.5) = (2n + 1) π/6, so t = 0.201 s, 0.866 s, 1.16 s, 1.74 s 6. x = Acos( ω t), v = – ω Asin( ω t), a = – ω 2 Asin( ω t) (a) sin( ω t) = +0.5, so ω t = (2n + 1) π/6, t = (2 n + 1) T/12 Find cos ( ω t) = +0.866, thus x = +0.866A (b) x = +A/2. Since cos( ω t) = +0.5, ω t = π(1, 2, 4, 5)/3 or t = T(1, 2, 4, 5)/6 7. (a) x = A sin(10t + φ ), v = 10A cos (10t + φ ) : a = 0.206 m. (1 + φ ) is in the 4th quadrant : (1 + φ ) = –1.33, So φ = –2.33 rad (b) x = 0.206 sin(10t – 2.33)m (c) (10t + φ ) is in the 2nd quadrant : sin(10t + φ ) = 0.971, (10t – 2.33) = 1.81 so t = 0.414 s. 8. x = Asin(4t + φ ), so v = 4Acos(4t + φ ), a = –16Asin(4t + φ ) At t = 0.15 s, –0.174 = 4Acos (0.6 + φ ), 0.877 = –16Asin(rt + φ ) Find tan(0.6 + φ ) = 1.26 and argument must be in 3rd quadrant. With (0.6 + φ ) = .4.04, φ = 3.44 rad. CHAPTER 15

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Then A = 0.07 m. x = 0.07 sin(4t + 3.44) m 9. (a) k = mg/x o = 30.6 N/m; ω = (k/m) 1/2 = 7.83 rad/s. Since x’ = A at t = 0, φ = π/2, so displacement from equilibrium x’ = 0.08sin(7.83t + π/2) m = 0.08 cos (7.83t)m. (b) If x = 0.1 m, then x’ = –0.06 , so cos(7.83t) = –0.75 and sin(7.83t) = 0.661, so v = ω A sin( ω t) = 0.414 m/s. a = – ω 2 x = +3.68 m/s 2 . 10. f 1 /f 2 = (m 2 /m 1 ) 1/2 , so 1.2/0.9 = [(m+50)/m] 1/2 Thus m = 64.3 g. f 1 = (k/m) 1/2 /2π, so k = 3.66 N/m. 11. ω = (k/m) 1/2 = 6.83 rad/s. x = A sin( ω t + φ ); v = ω A cos ( ω t + φ ), and a = – ω 2 x. (a) sin( ω t + φ ) = –1/3, so cos ( ω t + φ ) = ± 0.943 Thus v = ±0.773 m/s and a = +1.87 m/s 2 . (b) sin ( ω t + φ ) = +2/3. cos ( ω t + φ ) = 0.745 Thus v = ±0.611 m/s ; and a = –3.73 m/s 2 12. (a) F = k 1 x 1 = k 2 x 2 = (x 1 + x 2 )k eff ; F/k eff = F/k 1 + F/k 2 , so T = 2π(m/k e ff ) 1/2 = 2π[m(k 1 +k 2 )/k 1 k 2 ] 1/2 . (b) & (c) : F = k 1 (x + x o1 ) – k 2 (x o 2 – x) = (k 1 + k 2 ) = k e ff x 13. x = Rcos ( ω t) and y = R sin( ω t). Both indicate SHM. 14. ω = (k/m) 1/2 = 25.3 rad/s, A =0.2 m (a) K = 1/2 mv 2 =1/2 m[– ω Asin(2π/5)] 2 = 0.579 J U = 1/2 kx 2 = 1/2 k[Acos (2π/5)] 2 = 6.11 × 10 –2 J (b) cos 2 ( ω t) = 1/4, sin 2 ( ω t) = 3/4 K= 1/2 m ( ω A) 2 sin 2 ( ω t) =480 mJ U = 1/2 kA 2 cos 2 ( ω t) = 160 mJ (c) K = U when ω t = (2n +1) π/4 , thus t = (2n +1) T/8 when T = 2π/ ω =248 ms. 15. If K = U/2, then U = 2E/3 thsu x = ±(2/3) 1/2 A.
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Sol-Ch15 - CHAPTER 15 E xercises 1 cos 5/3 sin/6 2(a(20 t/4...

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