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.

Options

A. 0.05 m
B. 0.06 m
C. 0.01 m
D. 0.02 m

Answer

B

Explanation

In SHM, Total energy, `E=(1)/(2)momega^(2)A^(2)`
or `E=(1)/(2)m(2piupsilon)^(2)A^(2)" "(because omega=2piupsilon)`
`therefore A=(1)/(2piupsilon)sqrt((2E)/(m))`
Putting `E=K+U`, we get
`A=(1)/(2piupsilon)sqrt((2(K+U))/(m))=(1)/(2pi(25//pi))sqrt((2(0.5+0.4))/(0.2))=0.06m`