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

Correct answer: B. ω _ 2 ω _ 1 = n ^ 2 P _ 1max = ma ω _ 1 = b P _ 2max = mr ω _ 2 = R ω _ 1 ω _ 2 = 1 n ^ 2 ω _ 2 ω _ 1 = n ^ 2 E _ 1 = (1)/(2)m ω _ 1 ^ 2 a ^ 2 E _ 2 = (1)/(2)m ω _ 2 ^ 2 R ^ 2 E _ 1 E _ 2 = ω _ 1 ^ 2 ω _ 2 ^ 2 a ^ 2 R ^ 2 = ω _ 1 ^ 2 ω _ 2 ^ 2 n ^ 2 = ω _ 1 ^ 2 ω _ 2 ^ 2 ω _ 2 ω _ 1 E _ 1 E _ 2 = ω _ 1 ω _ 2 E _ 1 ω _ 1 = E _ 2 ω _ 2

JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Physics
Chapter: Oscillations

JEE Advanced 2015 · Paper 1 · Physics · Oscillations

Two independent harmonic oscillators of equal mass are oscillating...

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 $\dfrac{a}{b} = {n}^{2}$ and $\dfrac{a}{R} = n$ , then the correct equation(s) is (are)
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω _ 1 and

Options

A ${E}_{1}{\omega}_{1} = {E}_{2}{\omega}_{2}$
B $\dfrac{{\omega}_{2}}{{\omega}_{1}} = {n}^{2}$
C ${\omega}_{1}{\omega}_{2} = {n}^{2}$
D $\dfrac{{E}_{1}}{{\omega}_{1}} = \dfrac{{E}_{2}}{{\omega}_{2}}$

Correct answer

B. $\dfrac{{\omega}_{2}}{{\omega}_{1}} = {n}^{2}$

Step-by-step solution

{P}_{1max} = ma{\omega}_{1} = b

{P}_{2max} = mr{\omega}_{2} = R

\dfrac{{\omega}_{1}}{{\omega}_{2}} = \dfrac{1}{{n}^{2}}

\dfrac{{\omega}_{2}}{{\omega}_{1}} = {n}^{2}

{E}_{1} = \dfrac{1}{2}m{\omega}_{1}^{2}{a}^{2}

{E}_{2} = \dfrac{1}{2}m{\omega}_{2}^{2}{R}^{2}

\dfrac{{E}_{1}}{{E}_{2}} = {\omega}_{1}^{2}{\omega}_{2}^{2}\dfrac{{a}^{2}}{{R}^{2}} = {\omega}_{1}^{2}{\omega}_{2}^{2}{n}^{2} = {\omega}_{1}^{2}{\omega}_{2}^{2}\dfrac{{\omega}_{2}}{{\omega}_{1}}

\dfrac{{E}_{1}}{{E}_{2}} = \dfrac{{\omega}_{1}}{{\omega}_{2}}

\dfrac{{E}_{1}}{{\omega}_{1}} = \dfrac{{E}_{2}}{{\omega}_{2}}

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