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Ch14Word - Chapter14 1 (adistanceof2A)andbackagain.Thetotal...

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Chapter 14 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 ( ϖ 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 ( ϖ t  +  φ 29 , 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 ( ϖ t );     v  = –  v max  sin ( ϖ t );     a  = –  a max  cos ( ϖ t ). ( a ) We find  ϖ t  from v  = –  ! v max  = –  v max  sin ( ϖ t ), which gives  ϖ t  = 30 ° . Page  1
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Chapter 14 Thus the distance is x  =  x max  cos ( ϖ t ) =  x max  cos 30 °  =       0.866  x max   . ( b ) We find  ϖ t  from a  = –  !
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