(솔루션)물리학의 기초(북스힐) Solution - giambattista/McCarthy Richardson/Richardson 원저

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(솔루션)물리학의 기초(북스힐) Solution - giambattista/McCarthy Richardson/Richardson 원저에 대한 보고서 자료입니다.

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Chapter 10
ELASTICITY AND OSCILLATIONS
Conceptual Questions
1. Young’s modulus does not tell us which is stronger. Instead, it tells us which is more resistant to deformation for a
given stress. The ultimate strength would tell us which is stronger—i.e., which can withstand the greatest stress.
2. The pendulum should be lengthened to increase its period and slow down the clock.
3. When the block is struck, it initially begins to bend downward before actually breaking. Thus, the top of the block
is compressed, while the bottom is stretched. Since concrete has much less tensile strength than compressive
strength, it will break at the bottom first.
4. (a) Answer: 2F. Since stress is proportional to strain, the same force would produce the same strain on the bar of
half the length. However, strain is a relative change in length, defined as ΔL/L. Therefore, the change in
length would be half as much. To compress the bar by the same amount, then, would require a force twice as
great.
(b) Answer: F/4. To compress the bar by the same amount, with the same length as before, would require the
same stress. Stress is defined as F/A. With half the radius, the area is reduced to 1/4 of its initial value.
Therefore, to produce the same stress would require a force 1/4 as great as before.
5. The compressive force experienced by the columns is greater at the bottom than at the top, because the bottom
must support the weight of the column itself in addition to whatever the column is holding up. By increasing the
cross-sectional area of the bottom of the column, the stress it experiences is reduced. Tapering columns so that
they are thicker at the base prevents the stress at the bottom from being too large.
6. Although the distance traveled by the mass during each cycle is proportional to the amplitude of the oscillation,
the maximum velocity of the mass is as well. If the mass has farther to go, for example, it travels correspondingly
faster. This is how the period of the mass-spring system can be independent of amplitude.
7. Yes, the motion of the saw blade is SHM. The Scotch yoke effectively makes the horizontal displacement of the
saw blade equal to the x-component of the position of the knob, which is moving in a circle. When an object
moves in uniform circular motion, its x- (or y-) component exhibits SHM.
8. For the mass and spring system, the period will remain 1 s, because the period depends only on the mass and the
spring constant. For a pendulum, the period depends on the length and the gravitational field strength. With a
stronger gravitational field, the period of the pendulum would be less than 1 s.
9. The tension in the bungee cord at the lowest point would be greater than the person’s weight, because there is an
upward acceleration. In fact, the tension would have its maximum value at the bottom, because that is where the
upward acceleration is the greatest.
10. The breaking point of a rope is determined by the maximum strain it can withstand. The strain is the ratio of the
change in the length of the rope to the original length—the maximum strain is therefore independent of the rope’s
length. The strain is directly proportional to the stress—defined as the force per unit area. Thus, ropes of varying
length that are otherwise identical require the same force to reach the breaking point. The actual distance the rope
stretches before breaking is greater for a longer rope—more work, and thus more energy, is required to break a
longer rope.

키워드

물리,   물리학,   일반물리학,   물리학의 기초,   일반물리학 및 실험,   솔루션,   해답지,   giambattista,   McCarthy Richardson,   Richardson