Question

Potential energy of a simple harmonic oscillator at its mean position is $0.4$ J. If its kinetic energy at a displacement half of its amplitude from mean position is $0.6$ J, its total energy is

• A. $1.0J$
• B. $1.2J$
• C. $1.4J$
• D. $1.6J$

Answer 1

The correct option is B $1.2J$

PE min=0.4 J
($\therefore$ At mean positions KE is$ma{x}^m$ and PE is $mi{n}^m$)

$K{E}_1=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$K{E}_2=\frac{1}{2}m{\omega }^2{A}^2$

$\frac{K{E}_1}{K{E}_2}=\frac{(A^2-\frac{A^2}{4})}{A^2}$

$\frac{0.6}{K{E}_2}=\frac{3}{4}$

$K{E}_2=0.8J$

$\therefore TE=P{E}_{min}+K{E}_{max}$

$=0.4+0.8$
=1.23