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).

Options

A. `0.05m`
B. `0.06m`
C. `0.01m`
D. none of these

Answer

B

Explanation

`E=(1)/(2)momega^2A^2=(1)/(2)m(2pif)^2A^2`
`A=(1)/(2piF)sqrt((2E)/(m))`
Putting `E=K+U`, we get
`A=(1)/(2pi((25)/(pi)))sqrt((2xx(0.5+0.4))/(0.2))=0.06m)`