hw1_soln - Homework 1 Physics 3C Sp 08 page 1 of 7 a) x =...

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Unformatted text preview: Homework 1 Physics 3C Sp 08 page 1 of 7 a) x = (4 cm) cos( /7) = 3.6 cm b) v = dx/dt = -4 cm/s sin(/7) =-1.74 cm/s c) a = dv/dt = -4 cm/s^2 cos(/7) = -3.6 cm/s^2 d) T= 2/ = 2/(1 1/s) = 6.28 s a) = 3 /s = 2f => f = 3/2 Hz and T = 1/f = 2/3 s b) A is just 3m (just an exercise in understanding what amplitude is... c) (as with b this is just an exercise in understanding what the phase constant is) Homework 1 Physics 3C Sp 08 page 2 of 7 Homework 1 Physics 3C Sp 08 page 3 of 7 = (k/m)1/2 f = /2 => T = 1/f = 2 (m/k)1/2 k = (4 2 m/T2) = 41 N/m Homework 1 Physics 3C Sp 08 page 4 of 7 Homework 1 Physics 3C Sp 08 page 5 of 7 a) Main hint here: "haflway" means x = A/2. E_tot = 1/2 k A2 and E = 1/2(m v2 + k x2) Solve for v by setting (mv2 + k x2) = k A2 b) Use = (k/m)1/2 and period T = 2/ c) a_max = A2 Energy is conserved in the block. Use A = 11.5 cm, k = 7N/m. Period in Cambridge: T_c = 2 (L_c/g_c)1/2 Period in Tokyo: T_t = 2 (L_t/g_t)1/2 We know that T_t = T_c and this implies that (L_t/g_t) = (L_c/g_c) or that g_c/g_t = L_c/L_t = 0.995/0.993 = 1.002 Homework 1 Physics 3C Sp 08 page 6 of 7 A = L _max = (5m)(7o)(2/3600) = 0.61m and = (g/L)^1/2 = [9.8 m/s2/(5m)]1/2 = 1.4 rad/sec a)v_max = A = (0.61 m)(1.4 /s) = 0.855 m/s b)a_max = A 2 = 1.197 m/s2 and angular_acceleration = a/L ang_acc_max = a_max/L = 1.197m/s2/(5 m) = 0.239 rad/s2 c)F_max = ma_max = 0.25 kg (1.197 m/s) = 0.299 N Homework 1 Physics 3C Sp 08 page 7 of 7 The amplitude is simply the coefficient that multiplies the cosine in the equation: x = Ae-(b/2m)t cos( t + ) That is, the amplitude is just Ae-(b/2m)t At t = 0, the amplitude is 16 degrees. At t=1000s, the amplitude is 4 degrees. They ask for b/2m. At t=0, the amplitude is just A. So A = 16 degrees. At t = 1000, the amplitude is 16e-(b/2m)1000 = 4 e-(b/2m)1000s = 1/4 -(b/2m)1000s = ln(1/4) = -1.39 (b/2m) = 0.00139 s-1 ...
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hw1_soln - Homework 1 Physics 3C Sp 08 page 1 of 7 a) x =...

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