A particle of mass 'm' moves on a horizontal smooth line AB of length 'a' such that when particle is at any general point P on the line two forces act on it. A force 
towards A and another force 
towards B.
(i) Find its time period when released from rest from mid-point of line AB.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
force on particle at point P
F = 
– 
F = 
(2a – 3x)
F = 
(x – 2a/3)
(ii) So this is equation of S.H.M (F = mꞷ 2 x) so particle perform S.H.M with mean position x – 2a/3 = 0
x – 2a/3 = 0 x = 2a/3 (from point A)
So ꞷ 2 = 
⇒ ꞷ = 
So Time period T = 2 π 
and amplitude = 2a/3 – a/2 = a/6
minimum distance from B = a – (2a/3 + a/6) = a/6
(iii)
at point P velocity of particle = 0
and force = 
(at point q)
acc. = 
⇒ 
= 
⇒ V = 
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