REAL TEST 4
SAMPLE FINAL EXAM FOR AU 12
 QUIZ 11  (ANSWERS)
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. 
CH.  11; REAL TEST 4
CH. MULTIPLE CHOICE TBA  - Exercise/problems   26( try 27, 29) , 30 ( try 29, 31), 32 (try 33),   35, 37, 45, 47  
TURN IN: 26,  30,  32,   35, 37, 45, 47  
* DISCUSSIONS PROVIDED. 
SELECTED DISCUSSIONS TO exercises/problems BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
Vibrations 
26. Looking at this graph, I'm reminded of a block attached to a horizontal spring,   releases from rest and moving   toward the origin (unreformed spring) toward the left, say. See  figure 11.15. At t = 0.5 seconds, the mass arrives at the origin at is moving at the maximum speed allowed by conservation of energy. 
(a) The amplitude is the maximum distance from the origin, value seen by inspection of the graph.
(b) The period is the time between successive peaks read from the graph
(c) f = 1/period
(d) w = 6.28*f
30. Clearly the object is undergoing maximum speed since x = 0.  The total mechanical energy =(1/2)*mv2 + (1/2)*kx2  = constant or (1/2)*m*vf2  + (1/2)*kxi2   =  (1/2)*mvi2 + (1/2)*kxi2. relating energy at initial (i)  and final (f)  positions. When x = 0, 1/2*mv2    is a maximum, which means the block moves the fastest either  left or right at the origin ( x = 0, where the spring is un-deformed.) .

(a) Now, you can always write the total mechanical  energy as (1/2)*kA2 , where A is the amplitude. Thus,
(1/2)*kA2 = (1/2)*mv2 + (1/2)*kx2  , for any value of x or v.
Set x = 0  and v = -12.0 m/s and solve for A.
(b) The magnitude of the maximum acceleration is given by m*a =  k*A, where a is the magnitude of the maximum acceleration. Note  Fx = -kx, where Fx is the x-component of force exerted by spring on block; the minus sign indicates  spring  force always points to the origin,  when stretched ( x > 0) or compressed (x< 0).    Solve for acceleration magnitude a.  
(c) See previous [part (b)]  discussion. 
32.  This much is known: The object is slowing down since the leftward acceleration is opposite the rightward motion 
( velocity) . The object is slowing down as it moves toward x = A, where it stops for a moment before turning around.
We want to find A. Use:   (1/2)*kA2 = (1/2)*mv2 + (1/2)*kx2  , for any value of x or v. In particular use v = 2.20 m/s and the value of x = 0.600 m. But you must first find k/m from the relations, |- k*x |= m*a, where a is the acceleration magnitude at x = 0.600 m. From this we get   k/m = a/| x |  = (8.40 m/s2)/0.600 m
 Now re-write the energy conservation equation: 
 (1/2)*(k/m)*A2 = (1/2)*v2 + (1/2)*(k/m)*x2 .    Find A. Finally find A - 0.600 m. 
35. 
For (a) , (b), (c) and (d), see  problem 26 discussion.

(e) k/m = w2  , where w = 6.28*f. Given m= 2.40 kg,  find k. 
37.  A = (1.80  - 1.06)m /2 and f = 1/period = (1/2.15) s-1  = 0.46512   s-1. 

(a) k/m = w2  , where w = 6.28*f.  Given m, find k.  
(b) Use:   (1/2)*kA2 = (1/2)*mv2 + (1/2)*kx2 , for any value of x or v. Set x = 0 and find | v |  = maximum speed.
(c) m*a = k*A. Find acceleration magnitude a.
45. Use equation 11.32. The time period from the extreme left to extreme right side is one-half a period so be careful. 
47. Set 1.60 s = 6.28*square root of L/g, where g = 9.8 m/s2 ; see  equation 11.32.  Square both sides and solve for L.  Then plug L into the formula (11.32) for the period when g = 3.71 m/s2 and solve for the pendulum period on Mars to  celebrate the recent,  successful Mars probe launch!