JEE Advanced 2015 · Paper 1 · Physics Q11

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PhysicsMultiple correct+4 / -2Question 11 of 60

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 $\dfrac ab=n^2$ and $\dfrac aR=n$ , then the correct equation(s) is(are)

Two phase-space plots of momentum p against position x. Left (energy E₁): an ellipse centred at the origin with semi-axis a along x and b along p. Right (energy E₂): a circle of radius R centred at the origin.

E_1\omega_1=E_2\omega_2

\dfrac{\omega_2}{\omega_1}=n^2

\omega_1\omega_2=n^2

\dfrac{E_1}{\omega_1}=\dfrac{E_2}{\omega_2}

Answer and worked solution

Official answer: B, D

Phase-space orbits. For SHM, $x$ ranges up to $A$ and $p$ up to $m\omega A$ , and $E=\frac{p_{max}^2}{2m}$ .

Oscillator 1: $A=a$ and $b=m\omega_1a$ , so $\omega_1=\dfrac b{ma}$ and $E_1=\dfrac{b^2}{2m}$ .
Oscillator 2: a circle of radius $R$ , so $R=m\omega_2R$ , giving $\omega_2=\dfrac1m$ and $E_2=\dfrac{R^2}{2m}$ .

Check each option.

(B) $\dfrac{\omega_2}{\omega_1}=\dfrac ab=n^2$ . True.
(D) $\dfrac{E_1}{\omega_1}=\dfrac{ab}2$ and $\dfrac{E_2}{\omega_2}=\dfrac{R^2}2$ . With $b=\dfrac a{n^2}$ and $R=\dfrac an$ , both equal $\dfrac{a^2}{2n^2}$ . True.
(A), (C) do not hold in general.

(B), (D)

Topics:Energy in SHM Simple harmonic motion

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JEE Advanced 2015 · Paper 1, Physics question 11