The velocity of a particle in `S.H.M.` at positions `x_(1)` and `x_2` are `v_(1)` and `v_(2)` respectively. Determine value of time period and amplitude.
Text Solution
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`v = omegasqrt(A^(2)-x^(2)) rArr v^(2) = omega^(2)(A^(2) - x^(2))` At position `x_(1), v_(1)^(2) = omega^(2)(A^(2) - x_(1)^(2))`……(i) At position `x_(2) , v_(2)^(2) = omega^(2)(A^(2) - x_(2)^(2))` …..(ii) Subtracting `(ii)` from `(i)` from `v_(1)^(2) - v_(1)^(2) = omega^(2)(x_(2)^(2) - x_(1)^(2)) rArr omega = sqrt((v_(1)^(2) - v_(2)^(2))/(x_(2)^(2) - x_(1)^(2)))` Time period `t = (2pi)/(omega) rArr T = 2pisqrt((x_(2)^(2) - x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))` Dividing `(i)` by `(ii) (v_(1)^(2))/(v_(2)^(2)) = (A^(2) - x_(1)^(2))/(A^(2) - x_(2)^(2)) rArr v_(1)^(2)A^(2) - v_(1)^(2) xx x_(2)^(2) = v_(2)^(2) A^(2) - v_(2)^(2)x_(1)^(2)` So `A^(2) (v_(1)^(2) - v_(2)^(2)) = v_(1)^(2)x_(2)^(2) - v_(2)^(2)x_(1)^(2) rArr A = sqrt((v_(1)^(2)x_(2)^(2) - v_(2)^(2)x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))`
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