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 the 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)

$I^{st}$ harmonic oscillator.
$\frac{P^2}{b^2}+\frac{x^2}{a^2}=1$
$P^2=b^2\left(1-\frac{x^2}{a^2}\right)$
$P=\frac{b}{a}\sqrt{a^2-x^2}$
$\Rightarrow v=\frac{P}{m}$
$v=\frac{b}{am}\sqrt{a^2-x^2}$
Comparing $v=\omega \sqrt{A^2-x^2}$
${\omega }_1=\frac{b}{am},A_1=a$
$\&E_1=\frac{1}{2}m{\omega }_1^2{A}_1^2$
$I{I}^{nd}$ harmonic oscillation
$P^2+x^2=R^2$
$P=\sqrt{R^2-x^2}$
$v=\frac{P}{m}$
$v=\frac{1}{m}\sqrt{R^2-x^2}\left(v=\omega \sqrt{A^2-x^2}\right)$
Comparing${\omega }_2=\frac{1}{m},A_2=R_1$
$E_2=\frac{1}{2}m{\omega }_2^2{A}_2^2$
$\left(1\right)\frac{{\omega }_2}{{\omega }_1}=\frac{\frac{1}{m}}{\frac{b}{am}}$
$\Rightarrow \frac{{\omega }_2}{{\omega }_1}=\frac{a}{b}$ (given $\frac{a}{b}=n^2$)
$\frac{{\omega }_2}{{\omega }_1}=n^2$
$\left(2\right)\frac{E_1}{{\omega }_1}=\frac{1}{2}m{\omega }_1{A}_1^2$
$\Rightarrow \frac{E_1}{{\omega }_1}=\frac{1}{2}m\left(\frac{b}{am}\right){\left(a\right)}^2$
$\frac{E_1}{{\omega }_1}=\frac{1}{2}ab$
$\&\frac{E_2}{{\omega }_2}=\frac{1}{2}m\left({\omega }_2\right)A_2^2$
$\frac{E_2}{{\omega }^2}=\frac{1}{2}m\left(\frac{1}{m}\right){\left(R\right)}^2$
$\frac{E_2}{{\omega }_2}=\frac{R^2}{2}$
the value of ab $=\frac{a^2}{n^2}\left(\frac{a}{b}=n^2\right)$
$\&R^2=\frac{a}{n^2}\left(\frac{a}{R}=n\right)$
So$\frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$