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1) Define x=0 as the mass location when the spring
is relaxed (the mass is removed or is supported "by hand"), and take x as
positive upwards. Note this makes the elastic PE easy to evaluate as
simply .5*k*x^2 - the presence of other forces acting on the mass does not
change this. For convenience, also use the same x=0 point as the zero of
the gravitational PE.
2) The rest (equilibrium) position with the mass released (to
the earth's attraction) is x = a.
3) Energy balance at two arbitrary positions, x1 and x2 , in
the motion yields:
.5*k*x1^2 + m*g*x1 + .5*m*v1^2 = .5*k*x2^2 + m*g*x2 +
.5*m*v2^2
Given:
G1) With m=1 kg, a = -
.05m => k = -mg/a =9.8/.05 = 196N/m
G2) The suspended mass is released from rest at x =
-.25m
Apply Eq 3), letting x1 be the coordinate of the midpoint of
the oscillation, and x2 be the position of release from rest:
Numerically: x1 = a = -.05m; x2 = -.25m ; v2 = 0 (Also
remember k = -mg/a = 196N/m)
Eq 3) and algebra then yield v1 = 2.8m/sec
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