Question

If two bodies of same mass are executing S.H.M. with frequencies in the ratio $1:2$ and amplitudes in the ratio $2:3$ then the ratio of their total energies is

• A. $1:3$
• B. $1:9$
• C. $1:4$
• D. $1:6$

Answer 1

The correct option is B $1:9$

Total Energy =PE+KE
$PE=\frac{1}{2}m{\omega }^2{x}^2$
$KE=\frac{1}{2}m{\omega }^2(A^2-X^2)$
$T_2=PE+KE$
$=\frac{1}{2}m{\omega }^2{A}^2$
$m_1=m_2=m$
$f_1=f$
$f_2=2f$
$\therefore {\omega }_1=\omega (\therefore \omega =2\pi f)$
${\omega }_2=2\omega$
$A_1=2A$
$A_2=3A$
$\frac{T{E}_1}{T{E}_2}=\frac{\frac{1}{2}{m}_1{\omega }_1^2{A}_1^2}{\frac{1}{2}{m}_2{\omega }_2^2{A}_2^2}$
$=\frac{1}{9}$