A hanging spring stretches by 30.0 cm when an object of mass 490 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 20.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later?

(b) Find the distance traveled by the vibrating object in part (a).
(c) Another hanging spring stretches by 30.5 cm when an object of mass 480 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 20.0 cm and released from rest to oscillate without friction. Find its position x at a time 84.4 s later.
(d) Find the distance traveled by the vibrating object in part (c).
(e) Why are the answers to the parts (a) and (c) different by such a large percentage when the data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close?

To solve this problem, we need to use the concepts of simple harmonic motion and the equations of motion for a mass-spring system.

(a) To find the position of the object at a moment 84.4 s later, we need to determine the amplitude and phase constant of the object's motion. The amplitude can be calculated from the initial displacement when the object is released, which is the sum of the initial stretching and the additional displacement: 30.0 cm + 20.0 cm = 50.0 cm. The phase constant can be determined from the initial velocity, which is zero since the object is released from rest. Therefore, the position of the object at 84.4 s later can be found using the equation of motion for simple harmonic motion:
x(t) = A * cos(ωt + φ)

where x(t) is the position at time t, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

(b) To find the distance traveled by the vibrating object in part (a), we need to determine the total distance traveled during one complete cycle of the motion. Since simple harmonic motion is periodic, the object will return to the same position after one complete cycle. The total distance traveled in one cycle is given by 2 times the amplitude. Therefore, the distance traveled by the vibrating object in part (a) is 2 * 50.0 cm = 100.0 cm.

(c) For the second hanging spring with a different mass and initial displacement, we can follow a similar approach. The amplitude is now calculated as the sum of the initial stretching and the additional displacement: 30.5 cm + 20.0 cm = 50.5 cm. Again, the phase constant is zero since the object is released from rest. Using the equation of motion for simple harmonic motion, we can determine the position of the object at 84.4 s later.

(d) To find the distance traveled by the vibrating object in part (c), we again need to determine the total distance traveled during one complete cycle of the motion. Using the amplitude found in part (c), the distance traveled is 2 * 50.5 cm = 101.0 cm.

(e) The answers to parts (a) and (c) differ by a large percentage because the initial displacements are very close, but the amplitude values used in the calculations are slightly different due to the differences in the initial stretching of the springs. Since the distance traveled during one cycle is directly proportional to the amplitude, even a small difference in the amplitude can result in a significant difference in the total distance traveled. On the other hand, the answers to parts (b) and (d) are relatively close because they depend only on the amplitude, which is almost the same in both cases.