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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 19:41:38 IST
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a positively charged thin metal ring of radius R is fixed in the X-Y plane with its centre at the origin O.A negatively charged particle P is released fron rest at the point (0,0,z0) where z0>0. then the motion of P is-
a) periodic,for all values of z0 satisfying 0<z0<infinity
b) simple harmonic,for all values of z0 satisfying 0<z0<=R
c) approximately simple harmonic ,provided z0<<R
d) such that P crosses O and continues to move along the negative Z-axis towards z= -infinity
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 19:55:13 IST
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a and c are the right answers.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:02:04 IST
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plz solve it
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how is it an SHM
k*qx/(a^2+x^2)^3/2
so force on a particle of say -Q is given by
-k*qQx(a^2+x^2)^3/2
now if x<<a then force = -kqQx/a^3 which is an SHM
so c) is true
but for x------>inf force becomes inverse square so it is not SHM
so only c) is true
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:28:32 IST
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If R is the radius of the ring and z is the distance of a particle from the ring and on the axis of the ring then force on the particle is given by KQz/(z^2+R^2)^3/2.
NOW In the first option it perform a periodic motion because ring is attracting the particle towards itself and due to inertia it will cross the origin and same in the backward motion.
In the third option ,if zis much less than R then we can write the above equation as KQz/R^3.now it seems to be proprtional to z ,hence perform S.H.M
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:34:59 IST
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but jinu as x becomes large it becomes an inverse square force so how can it be periodic as attractive force decreases with distance ?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:38:56 IST
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but the answer is both a and c
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:44:36 IST
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check the option yaar it is less than infinity and not equal to infinity.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 20:47:29 IST
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in the options it is given as z= -infinity
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 21:01:34 IST
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I think both options a and c. jinu.coolboy has shown that it is SHM for displacement of particle << radius of ring.
Option a is right as force is always towards the centre of the ring and of equal magnitude for same distance from centre of ring so it is periodic.
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