Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_{1}$ and $\omega_{2}$ and have total energies $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$, respectively. The variations of their momenta p with positions x are shown in figures. If $\frac{\mathrm{a}}{\mathrm{b}}=\mathrm{n}^{2}$ and $\frac{\mathrm{a}}{\mathrm{R}}=\mathrm{n}$, then the correct equation(s) is (are) :

(B) $\frac{\omega_{2}}{\omega_{1}}=\mathrm{n}^{2}$; (D) $\frac{\mathrm{E}_{1}}{\omega_{1}}=\frac{\mathrm{E}_{2}}{\omega_{2}}$

(A) $\mathrm{E}_{1} \omega_{1}=\mathrm{E}_{2} \omega_{2}$

(C) $\omega_{1} \omega_{2}=\mathrm{n}^{2}$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Elasticity & SHM

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 figures. If

\frac{a}{b} = n^{2}

and

\frac{a}{R} = n

, then the correct equation(s) is (are) :

Answer: (B), (D)

For a harmonic oscillator, the phase space ellipse has semi-axes related by

p_{ma x} = mω x_{ma x}

. From the first figure,

b = ma ω_{1}

, so

a / b = 1/ (m ω_{1}) = n^{2}

, giving

ω_{1} = 1/ (m n^{2})

.

From the second figure,

R = ma ω_{2}

. Since

a / R = n

, we have

R = a / n = ma ω_{2}

, so

ω_{2} = 1/ (mn)

.

Therefore

ω_{2} / ω_{1} = n^{2}

, confirming (B).

The total energy is

E = (1/2) m ω^{2} A^{2}

where A is amplitude. For both oscillators with amplitude a:

E_{1} / ω_{1} = (1/2) m a^{2} ω_{1}

and

E_{2} / ω_{2} = (1/2) m a^{2} ω_{2}

. Since both have the same amplitude a and mass m,

E_{1} / ω_{1} = E_{2} / ω_{2}

, confirming (D).

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options