Ch14Word - Chapter 14 CHAPTER 14 Oscillations 1 In one period the particle will travel from one extreme position to the other(a distance of 2A and

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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 ( t + φ ) , so x ( t = 0) = A cos . ( a ) – A = A cos , which gives cos = – 1, so = p (or –p) . ( b ) 0 = A cos , which gives cos = 0, so = p/2 (or 3p/2) . ( c ) A = A cos , which gives cos = + 1, so = 0 . ( d ) ! A = A cos , which gives cos = ! , so = p/3 (or –p/3) . ( e ) – A /2 = A cos , which gives cos = – ! , so = 2p/3 (or 4p/3) . ( f ) A /v2 = A cos , which gives cos = 1/v2, so = p/4 (or –p/4) . 7. Because the mass is released at the maximum displacement, we have x = x max cos ( t ); v = – v max sin ( t ); a = – a max cos ( t ). ( a ) We find t from v = – ! v max = – v max sin ( t ), which gives t = 30°. Thus the distance is x = x max cos ( t ) = x max cos 30° = 0.866 x max . ( b ) We find t from a = – ! a max = – a max cos ( t ), which gives t = 60°. Thus the distance is x = x max cos ( t ) = x max cos 60° = 0.500 x max .
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Chapter 14 Page 2 8. ( a ) We find the effective spring constant from the frequency: f 1 = ( k / m 1 ) 1/2 /2p; 2.5 Hz = [ k /(0.050 kg)] 1/2 /2p, which gives k = 12 N/m . ( b ) Because the size and shape are the same, the spring constant, which is determined by the buoyant force, will be the same. The new frequency of vibration will be f 2 = ( k / m 2 ) 1/2 /2p = [(12 N/m)/(0.25 kg)] 1/2 /2p = 1.1 Hz . 9. For a general displacement, we have x = A cos ( ω t + φ ); v = – A sin ( t + ). We find t + from v = – ! A = – A sin ( t + ), which gives t + = 30°. Thus the displacement is x = A cos ( t + ) = A cos (30°) = 0.866 A . 10. The dependence of the frequency on the mass is f = ( k / m ) 1/2 /2p. Because the spring constant does not change, we have f 2 / f = ( m / m 2 ) 1/2 ; (0.48 Hz)/(0.88 Hz) = [ m /( m + 1.25 kg)] 1/2 , which gives m = 0.53 kg . 11. In the equilibrium position, the net force is zero. When the mass is pulled down a distance x , the net restoring force is the sum of the additional forces from the springs, so we have F net = ? F 2 + ? F 1 = – k 2 x k 1 x = – ( k 1 + k 2 ) x , which gives an effective force constant of k 1 + k 2 .
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch14Word - Chapter 14 CHAPTER 14 Oscillations 1 In one period the particle will travel from one extreme position to the other(a distance of 2A and

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