Question

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ${\omega }_1$ and ${\omega }_2$ and have total energies $E_1$ and $E_2$, respectively. The variations of their momenta p with positions x are shown in figures. If $\frac{a}{b}=n^2$ and $\frac{a}{R}=n$, then the correct equation(s) is (are)

• A. $E_1{\omega }_1=E_2{\omega }_2$
• B. $\frac{{\omega }_2}{{\omega }_1}=n^2$
• C. ${\omega }_1{\omega }_2=n^2$
• D. $\frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$

Answer 1

Answer: (B), (D)

$\frac{{\omega }_2}{{\omega }_1}=n^2$
$\frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$

First oscillator:
$\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2=E$
Comparing this with equation of ellipse having semi-major and semi-minor axes and b respectively,
$\frac{x^2}{a^2}+\frac{p^2}{b^2}=1$
$b^2=2mE$
$a^2=\frac{2E}{m{\omega }_1^2}$
$\Rightarrow \frac{a}{b}=\frac{1}{m{\omega }_1}$ ....(1)
2nd oscillator:
$\frac{x^2}{R^2}+\frac{p^2}{R^2}=1$
$\Rightarrow \frac{R}{R}=\frac{1}{{\omega }_2}$ ....(2)
Dividing (1) by (2), $\frac{a}{b}=\frac{{\omega }_2}{{\omega }_1}$
$\Rightarrow n^2=\frac{{\omega }_2}{{\omega }_1}$
$E_1=\frac{1}{2}m{\omega }_1^2{a}^2$
$E_2=\frac{1}{2}m{\omega }_2^2{R}^2$
$\frac{E_1}{E_2}=\frac{{\omega }_1^2}{{\omega }_2^2}\frac{a^2}{R^2}=\frac{1}{n^4}{n}^2$
$\Rightarrow \frac{E_1}{E_2}=\frac{1}{n^2}=\frac{{\omega }_1}{{\omega }_2}$
$\Rightarrow \frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$