A particle is executing simple harmonic motion in a conservative force field. The total energy of simple harmonic motion is given by `E=ax^(2)+bv^(2)`where ‘x’ is the displacement from mean position x = 0 and v is the velocity of the particle at x then choose the INCORRECT statements.{Potential energy at mean position is assumed to be zero}

Options

A. amplitude of S.H.M. is `sqrt(E/a)`
B. Maximum velocity of the particle during S.H.M. is `sqrt(E/b)`
C. Time period of motion is `2pi sqrt(b/a)`
D. displacement of the particle is proportional to the velocity of the particle.

Explanation

Amplitude is obtained for v=0
`:. A=sqrt(E/a)`
maximum velocity is obtained for x=0
`V_(max)=sqrt(E/b) V_(max)=A omega`
`omega=(sqrt(E/b))/(sqrt(E/a))=sqrt(a/b)`
`T=(2pi)/(omega)=2pisqrt(b/a)`
Alternative
`E=1/2 mv^(2)+1/2 kx^(2)`
`b=m/2,a=k/2`
`omega=sqrt(k/m)=sqrt(a/b)`
`E=1/2 mv_(max)^(2)rArr V_(max)=sqrt(E/b)`
`E=1/2 lA^(2) A=sqrt(E/a)`