JEE Advanced · Physics · 15. Oscillations

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)

A $E_{1} ω_{1} = E_{2} ω_{2}$
B $\frac{ω_{2}}{ω_{1}} = n^{2}$
C $ω_{1} ω_{2} = n^{2}$
D $\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

Step-by-step Solution

Detailed explanation

P_{1 m a x} = m a ω_{1} = b

P_{2 m a x} = m r ω_{2} = R

\frac{ω_{1}}{ω_{2}} = \frac{1}{n^{2}}

\frac{ω_{2}}{ω_{1}} = n^{2}

E_{1} = \frac{1}{2} m ω_{1}^{2} a^{2}

E_{2} = \frac{1}{2} m ω_{2}^{2} R^{2}

\frac{E_{1}}{E_{2}} = \frac{ω_{1}^{2}}{ω_{2}^{2}} \frac{a^{2}}{R^{2}} = \frac{ω_{1}^{2}}{ω_{2}^{2}} n^{2} = \frac{ω_{1}^{2}}{ω_{2}^{2}} \frac{ω_{2}}{ω_{1}}

\frac{E_{1}}{E_{2}} = \frac{ω_{1}}{ω_{2}}

\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

List I describes four systems, each with two particles

A

and

B

in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in

m s^{- 1}

) at time

t = \frac{π}{3} s

	List-I		List-II
(I)	$A$ and $B$ are moving on a horizontal circle of radius $1 m$ with uniform angular speed $ω = 1 rad s^{- 1}$ . The initial angular positions of $A$ and $B$ at time $t = 0$ are $θ = 0$ and $θ = \frac{π}{2}$ respectively.	(P)	$\frac{\sqrt{3} + 1}{2}$
(II)	Projectiles $A$ and $B$ are fired (in the same vertical plane) at $t = 0$ and $t = 0.1 s$ respectively, with the same speed $v = \frac{5 π}{\sqrt{2}} m s^{- 1}$ and at $45 °$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide, $(g = 10 m s^{- 2})$ .	(Q)	$\frac{(\sqrt{3} - 1)}{\sqrt{2}}$
(III)	Two harmonic oscillators $A$ and $B$ moving in the $x$ direction according to $x_{A} = x_{0} \sin \frac{t}{t_{0}}$ and $x_{B} = x_{0} \sin (\frac{t}{t_{0}} + \frac{π}{2})$ respectively, starting from $t = 0$ . Take $x_{0} = 1 m, t_{0} = 1 s$ .	(R)	$\sqrt{10}$
(IV)	Particle $A$ is rotating in a horizontal circular path of radius $1 m$ on the $x y$ plane, with constant angular speed $ω = 1 rad s^{- 1}$ . Particle $B$ is moving up at a constant speed $3 m s^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.)	(S)	$\sqrt{2}$
		(T)	$\sqrt{25 π^{2} + 1}$

Which one of the following options is correct?

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From JEE Advanced

A series LCR circuit is connected to a

45 \sin (ω t)

Volt source. The resonant angular frequency of the circuit is

10^{5} rad s^{- 1}

and current amplitude at resonance is

I_{0}

. When the angular frequency of the source is

ω = 8 \times 10^{4} rad s^{- 1}

, the current amplitude in the circuit is

0.05 I_{0}

. If

L = 50 mH

, match each entry in List-

I

with an appropriate value from List-

II

and choose the correct option.

	List- $I$		List- $II$
$P$	$I_{0}$ in $mA$	$1$	$44.4$
$Q$	The quality factor of the circuit	$2$	$18$
$R$	The bandwidth of the circuit in $rad s^{- 1}$	$3$	$400$
$S$	The peak power dissipated at resonance in Watt	$4$	$2250$
		$5$	$500$

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