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A particle executes linear some with an amplitude if 2 cm.when the particle is at 1cm from the mean position the magnitude of its velocity is equal to that of its acceleration.Then its time period is?Ans) velocity at distance x in shm is given as v= w sqrt(A2-x2)so A = 2 x =1v = w sqrt(4-1) = wsqrt3now acc at distance x is a= w2 x = w2as mag a = mag vw2 = wsqrt3w= sqrt32pi/T= sqrt3T = 2pi/sqrt3
q+q''=0
neutral point of bar magnet
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