Asked by Daphine
A particle P of mass m=1 moves on the x-axis under the force field
F=36/(x^3) - 9/(x^2) (x>0).
(a)Initially P is projected from the point x=4 with speed 0.5. Show that P oscillates between two extremes points and find the period of the motion.
(b)Show that there is a single equilibrium position for P and that it is stable. Find the period of small oscillations about this point.
F=36/(x^3) - 9/(x^2) (x>0).
(a)Initially P is projected from the point x=4 with speed 0.5. Show that P oscillates between two extremes points and find the period of the motion.
(b)Show that there is a single equilibrium position for P and that it is stable. Find the period of small oscillations about this point.
Answers
Answered by
Mgraph
F=ma where a is the acceleration of
the particle P.
a=dv/dt where v is the velocity.
dv/dt=(dv/dx)(dx/dt)=(dv/dx)v
(dv/dx)v=36/(x^3)-9/(x^2)
(v)dv=(36/(x^3)-9/(x^2))dx
Integrating we get
v^2/2=-18/(x^2)+9/x+C
If t=0 then x=4 and v=0.5
C=-1
v^2/2=(6-x)(x-3)/(x^2) therefore
3<=x<=6 =>P oscillates between x=3 and
x=6
the particle P.
a=dv/dt where v is the velocity.
dv/dt=(dv/dx)(dx/dt)=(dv/dx)v
(dv/dx)v=36/(x^3)-9/(x^2)
(v)dv=(36/(x^3)-9/(x^2))dx
Integrating we get
v^2/2=-18/(x^2)+9/x+C
If t=0 then x=4 and v=0.5
C=-1
v^2/2=(6-x)(x-3)/(x^2) therefore
3<=x<=6 =>P oscillates between x=3 and
x=6
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