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# Ch14Word - Chapter 14 CHAPTER 14 Oscillations 1 In one...

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Chapter 14 Page 1 CHAPTER 14 – Oscillations 1. In one period the particle will travel from one extreme position to the other (a distance of 2 A ) and back again. The total distance traveled is d = 4 A = 4(0.15 m) = 0.60 m . 2. ( a ) We find the spring constant from the elongation caused by the weight: k = mg /? x = (3.7 kg)(9.80 m/s 2 )/(0.028 m) = 1.30 × 10 3 N/m . ( b ) Because the fish will oscillate about the equilibrium position, the amplitude will be the distance the fish was pulled down from equilibrium: A = 2.5 cm . The frequency of vibration will be f = ( k / m ) 1/2 /2p = [(1.30 × 10 3 N/m)/(3.7 kg)] 1/2 /2p = 3.0 Hz . 3. We find the spring constant from the compression caused by the increased weight: k = mg / x = (80 kg)(9.80 m/s 2 )/(0.0140 m) = 5.60 × 10 4 N/m. The frequency of vibration will be f = ( k / m ) 1/2 /2p = [(5.60 × 10 4 N/m)/(1080 kg)] 1/2 /2p = 1.15 Hz . 4. ( a ) Because the motion starts at the maximum extension, we have x = A cos ( ω t ) = A cos (2p t / T ) = (8.8 cm) cos [2p t /(0.75 s)] . ( b ) At t = 1.8 s we get x = (8.8 cm) cos [2p(1.8 s)/(0.75 s)] = – 7.1 cm . 5. ( a ) We find the effective spring constant from the frequency: f 1 = ( k / m 1 ) 1/2 /2p; 10 Hz = [ k /(0.60 × 10 –3 kg)] 1/2 /2p, which gives k = 2.4 N/m . ( b ) The new frequency of vibration will be f 2 = ( k / m 2 ) 1/2 /2p = [(2.37 N/m)/(0.40 × 10 –3 kg)] 1/2 /2p = 12 Hz . 6. The general expression for the displacement is x = A cos ( ω t + φ ) , so x ( t = 0) = A cos φ . ( a ) – A = A cos φ , which gives cos φ = – 1, so φ = p (or –p) . ( b ) 0 = A cos φ , which gives cos φ = 0, so φ = p/2 (or 3p/2) . ( c ) A = A cos φ , which gives cos φ = + 1, so φ = 0 . ( d ) ! A = A cos φ , which gives cos φ = ! , so φ = p/3 (or –p/3) . ( e ) – A /2 = A cos φ , which gives cos φ = – ! , so φ = 2p/3 (or 4p/3) . ( f ) A /v2 = A cos φ , which gives cos φ = 1/v2, so φ = p/4 (or –p/4) . 7. Because the mass is released at the maximum displacement, we have x = x max cos ( ω t ); v = – v max sin ( ω t ); a = – a max cos ( ω t ). ( a ) We find ω t from v = – ! v max = – v max sin ( ω t ), which gives ω t = 30°. Thus the distance is x = x max cos ( ω t ) = x max cos 30° = 0.866 x max . ( b ) We find ω t from a = – ! a max = – a max cos ( ω t ), which gives ω t = 60°. Thus the distance is x = x max cos ( ω t ) = x max cos 60° = 0.500 x max .

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