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)*mv^{2} +
(1/2)*kx^{2 } = constant or (1/2)*m*v_{f}^{2}
+ (1/2)*kx_{i}^{2 } = (1/2)*mv_{i}^{2}
+ (1/2)*kx_{i}^{2}. relating energy at initial (i)
and final (f) positions. When x = 0, 1/2*mv^{2 }
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)*kA^{2} , where A is the amplitude. Thus, |
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)*kA^{2} = (1/2)*mv^{2} + (1/2)*kx^{2 } , 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/s^{2})/0.600 m Now re-write the energy conservation equation: (1/2)*(k/m)*A^{2} = (1/2)*v^{2} + (1/2)*(k/m)*x^{2 . }Find A. Finally find A - 0.600 m. |
35. For (a) , (b), (c) and (d), see problem 26 discussion. (e) k/m = w^{2} , 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 = w^{2} , where w = 6.28*f. Given m,
find k. |
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/s^{2} ; 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/s^{2} and solve for the pendulum period on Mars to celebrate the recent, successful Mars probe launch! |