You have a spring hanging vertically. You add a mass to the end and let it...

You have a spring hanging vertically. You add a mass to the end and let it stretch to find its new equilibrium position which is 3.17 cm below the unstretched position. You now raise the mass (compress the mass-spring system) to its original unstretched position. Now you release the mass. How far from this starting position does it travel downward until it stops and changes direction?

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Expert Solution

When unstretched, the position of the spring is shown as point A.

After hanging a mass to the spring's end, the spring stretches and the new equilibrium position of the spring-mass is shown as point B. Point B is at distance 3.17cm from point A. Since point B is equilibrium position of the system, any displacement given to the mass will result in an oscillatory motion about this point.

The spring is compressed to point A and released from rest. The mass will undergo up and down oscillations about equilibrium point B. The amplitude of oscillation is the distance between points A and B, which is 3.17 cm. The mass will pass point B and travel 3,17cm downward to point C. It will momentarily stop at point C, and move up again.

The distance traveled by the mass starting from point A until it stops is the distance between point A and C.

genius_generous answered 1 year ago

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