InterviewSolution
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InterviewSolution
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A partical of mass `0.2 kg` executes simple harmonic motion along a path of length `0.2 m` at the rate of `600`oscillations per minute. Assum at `t = 0`. The partical start SHM in positive direction. Find the kinetic potential energies in zoules when the displacement is `x = A//2` where, A stands for the amplitude. |
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Answer» Given that `f = 600` oscillations//min, `m = 0.2 kg` `A = "Amplitude" = 1//2 xx "Length of path" = 0.1 m = (1)/(10) m` As `v = (1)/(T) = (600)/(60) = 10 Hz impliesomega = 2 pi T = 20 pi rad//s` The magnetitude velocity of a partical performing SHM is given as `v = omega sqrt(A^(2) - x^(2))`. At `x = A//2 = (1)/(20) m`, the velocity at this position `v_(x = A//2) = 20 pi sqrt (((1)/(10)^(2) = (1)/(20)^(2))) = sqrt 3 pi m//s` Hence, kinetic energy at this position `k_(x = A//2) = (1)/(2) mv_(x = A//2) ^(2) = (1)/(2) (0.2) (sqrt 3 pi)^(2) = (3 pi^(2))/(10) J` We can write potential energy at any position `x` as `U = (1)/(2)kx^(2) = (1)/(2) (m omega^(2)) x^(2)` Hence, `U = (1)/(2) (0.2) (20 pi)^(2) ((1)/(20))^(2) = (pi^(2))/(10) J` |
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