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An object of mass 0.2 kg executes simple...

An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m.

Text Solution

Verified by Experts

The correct Answer is:
0.06
`x=0.04m`,`KE=0.5J` and `PE=0.4J`
Also `v=(25)/(pi)Hx` , Now, `TE=(1)/(2)momega^(2)a^(2)=(1)/(2)mxx4pi^(2)v^(2)a^(2)`
implies `0.9=(1)/(2)xx0.2xx4pi^(2)xx(25)/(pi)xx(25)/(pi)xxa^(2)` implies `a=(3)/(50)=0.06m`
Alternate Method:
`(1)/(2)mv^(2)=0.5implies(1)/(2)(0.2)v^(2)=0.5impliesv^(2)=5`
Now `v^(2)=omega^(2)(A^(2)-x^(2))` implies `5(2piv)^(2)[A^(2)-(0.04)^(2)]`
implies `5=4pi^(2)xx(25)/(pi)xx(25)/(pi)[A^(2)-(0.04)^(2)]` implies `A=0.06`.
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