(a) A hanging spring stretches by 32.0 cm when an object of mass 480 g is hung o

Question

(a) A hanging spring stretches by 32.0 cm when an object of mass 480 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 16.5 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 32.5 cm when an object of mass 470 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 16.5 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?

Does this circumstance reveal a fundamental difficulty in calculating the future? Yes or No?

Explanation / Answer

mg = kx

k = mg/x = 0.48 * 9.8 / 0.32 = 14.7 N/m

x = A cost

A = 0.155 m

= sqrt(k/m) = sqrt(14.7/0.48) = 5.53

at t = 84.4 s

x = 0.165cos(5.53 * 84.4) = 0.48 m

(b) period T = 2 / = 2 sqrt(m/k) = 2 sqrt(0.48/14.7) = 1.135 s

                84.4 / 1.135 = 74.3

the object finished 74.3 oscilations in 84.4 seconds.

distance travled = 74.3 * 2A = 70 * 2 * 0.165 = 23.1 m

(c) k = 0.48*9.8 / 0.325 = 14.5 N/m

          x = Acost

         A = 0.165 m

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