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

Answer

0.06

Explanation

`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`.