Q.
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $ω_1$ and $ω_2$ 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)
JEE AdvancedJEE Advanced 2015
Solution:
$I^{st}$ 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 $\quad v = \omega\sqrt{A^{2}-x^{2}}$
$\omega_{1} = \frac{b}{am}, A_{1} = a$
$\& \,E_{1} = \frac{1}{2} m \omega^{2}_{1} A^{2}_{1}$
$II^{nd}$ harmonic oscillation
$P^{2} + x^{2} = R^{2}$
$P = \sqrt{R^{2}-x^{2}}$
$v = \frac{P}{m}$
$v = \frac{1}{m} \sqrt{R^{2}-x^{2}}\quad\quad\left(v = \omega\sqrt{A^{2}-x^{2}}\right)$
Comparing$\quad\omega_{2} = \frac{1}{m}, A_{2} = R_{1}$
$E_{2} = \frac{1}{2} m \,\omega^{2}_{2} A^{2}_{2}$
$\left(1\right)\quad \frac{\omega_{2}}{\omega_{1}} = \frac{\frac{1}{m}}{\frac{b}{am}}$
$\Rightarrow \quad \frac{\omega_{2}}{\omega_{1}} = \frac{a}{b}\quad\quad$ (given $\frac{a}{b} = n^{2}$)
$\frac{\omega_{2}}{\omega_{1}} = n^{2}$
$\left(2\right)\quad \frac{E_{1}}{\omega_{1}} = \frac{1}{2} m\omega_{1} A^{2}_{1}$
$\Rightarrow \quad \frac{E_{1}}{\omega_{1}} = \frac{1}{2} m \left(\frac{b}{am}\right) \left(a\right)^{2}$
$\frac{E_{1}}{\omega_{1}} = \frac{1}{2} ab$
$\&\quad \frac{E_{2}}{\omega_{2}} = \frac{1}{2} m \left(\omega_{2}\right) A^{2}_{2}$
$\frac{E_{2}}{\omega^{2}} = \frac{1}{2} m \left(\frac{1}{m}\right) \left(R\right)^{2}$
$\frac{E_{2}}{\omega_{2}} = \frac{R^{2}}{2}$
the value of ab $= \frac{a^{2}}{n^{2}} \left(\frac{a}{b} = n^{2}\right)$
$\& \quad R^{2} = \frac{a}{n^{2}} \left(\frac{a}{R} = n\right)$
So$\quad \frac{E_{1}}{\omega_{1}} = \frac{E_{2}}{\omega_{2}}$
