Question

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.

Answer 1

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`