449_Physics ProblemsTechnical Physics

449_Physics ProblemsTechnical Physics - Chapter 15 P15.33...

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Chapter 15 451 P15.33 Referring to the sketch we have Fm g =− sin θ and tan = x R For small displacements, tan sin θθ and F mg R xk x Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion. Comparing to x ω 2 shows == k m g R . FIG. P15.33 P15.34 (a) T = total measured time 50 The measured periods are: Length, m Period, s L T a f a f 1 000 0 750 0 500 1 996 1 732 1 422 ... (b) T L g = 2 π so g L T = 4 2 2 The calculated values for g are: Period, s ms 2 T g af ej 1 996 1 732 1 422 991 987 976 4 3 2 1 0 0.25 0.5 0.75 1.0 L , m T 2 , s 2 FIG. P15.34 Thus, g ave 2 = 985 . this agrees with the accepted value of g = 980
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