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Obtain instantaneous velocity of a particle executing SHM. |
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Answer» SOLUTION :Intantaneous velocity of SHM particle is the rate of change of displacement for a time interval. Suppose the displacement of SHM particle at time t with amplitude A and angular speed `omega` is `x(t)= A cos (omega t+ phi)"""……"(1)` where `phi` is the initial phase DIFFERENTIATING EQUATION (1) respect to t gives the instantaneous velocity `therefore v(t)= (d[x(t)])/(dt) = (d)/(dt)[A cos (omega t+ phi)]` `therefore v(t)= -A omega sin (omega t+ phi)` but `sin(omega t +phi)= pm sqrt(1-cos^(2) (omega t+ phi))` `v(t)= (-A omega) (pm sqrt(1-cos^(2) (omega t+ phi)))` `= pm omega sqrt(A^(2)-A^(2) cos^(2)(omega t+ phi))` `= pm omega sqrt(A^(2)-x^(2)(t))` Generally `v= pm omega sqrt(A^(2) - x^(2))` Special Cases : (1) At mean position, x=0 Velocity `v= pm omega A` In +x-direction, v is positive and in -x-direction, v is negative. Hence, maximum velocity of SHM particle `v_("max")= A omega` (2) At extreme point, for velocity `x= |A|`, `v= pm omega sqrt(A^(2)-A^(2))""[therefore |A|^(2)= A^(2)]` `therefore v= 0` The velocity of SHM particle at extreme points is ZERO. |
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