When the system shown is slowly reased the previously unstretched spring is stre
ID: 1448349 • Letter: W
Question
When the system shown is slowly reased the previously unstretched spring is stretched a distance Xo from its equilibrium point. The mass M is then pulled down so as to be a distance X1 from its original position. (X1>Xo) When M is released the spring will shorten. How far will M be from it's orginally equilibrium point at the maximum compression point of the spring? (M>m). (this is the point where M stops going up)
The picture shows a pully with mass m hanging off on the left sitting on top of a spring and mass M hanging off the right side.
Explanation / Answer
The system undergoes simple harmonic motion about the equilibrium position
since it is pulled to a distance of x1 downward
amplitude will be X1
so the mass M goes up to the same distance X1 above the equilibrium position
if u have any doubts u can comment
simple logic mass M oscilates about equilibrium
since it has been pulled x1 dostance from equilibrium
it goes x1 distance above equilibrium
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