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A 20g partical is oscillating simple bar...

A `20g` partical is oscillating simple barmonically with a period of `2 second` and maximum kinetic energy `2 J`. The total machanical energy of the partical is zero , find
a Amplitude of oscillation
b. potential energy as a punction of displacement x relative to mean position.

Text Solution

Verified by Experts

a. Given `m = 20g = 20 xx 10^(-3) kg`
Time period , `T = 2 s and K_(max) = 2 J`
We havemaximum kinetic energy : `K_(max) = (1)/(2) m omega^(2) A^(2)`
Hence, `2 = (1)/(2) xx 20 xx 10^(-3) xx ((2 pi)/(2))^(2) . A^(2) implies A = (10sqrt2)/(pi) m`
b. We know total energy of a partical is of oscillation energy and minimum poitential energy.
`E_(total) = E_(oscillation) + U_(0)`
But oscillation energy of partical: ` E_(oscillation) = K_(max) = 2 J`
As potential energy in reation with displacement `x` from mean position is
`U (x) = (1)/(2) k x^(2) + U_(0) = (1)/(2) (m omega^(2)) x^(2) + U_(0)`
`= (1)/(2) (m omega^(2)) x^(2) - 2`
`(1)/(2) xx (20)/(1000) xx ((2 pi)/(2))^(2) x^(2) - 2`
`= (pi^(2))/(100)x^(2) - 2` (S.I. units)
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