The P.E. of a linear harmonic oscillator is 4J, when its displacment is equal to half its amplitude. The total energy of the particle is

Options

A. 8 J
B. 32 J
C. 16 J
D. 24 J

Answer

C

Explanation

`(1)/(2)Kx^(2)=4 therefore (1)/(2)K((A^(2))/(4))=4`
`therefore E=(1)/(2)KA^(2)=16`