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Equations of SHM A particle is subjected to two SHMs x1 = A1sin wt and x2 = A2 sin (wt + 180/4). The resultant SHM will have an amplitude of treat w as omega Answer is( A12 +A22+ (2)1/2A1 A2) 1/2 how A particle moves according to the law x=a cos pi.t/2. The distance covered by it in the time interval between t =0 to t=3 sec is 1)2a 2)3a 3)4a 4)a which is correct answers please.
x=Asinωt (i)y=Asin(2ωt+π/2) =Acos2ωt =A(1−2sin2ωt)From Eq. (i)sinωt=Ax∴ y=A(1−A22x2) Thanks and regards
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