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?

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 2 years ago

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