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) [2015]

Question

Smart Achievers · Accepted Answer

(2, 4)

$For first harmonic oscillator,$

$Mass = m$

$Angular frequency = ω_{1}$

$x$

$Total energy = E_{1}$

$Maximum momentum, p_{\max} = b$

$E_{1} = \frac{1}{2} m ω_{1}^{2} a^{2}$ ...(i)

$p_{\max} = m v_{\max} = m a ω_{1} \Rightarrow b = m a ω_{1}$

$\frac{a}{b} = \frac{1}{m ω_{1}}$ ...(ii)

$For second harmonic oscillator,$

$Mass = m$

$Angular frequency = ω_{2}$

$Amplitude = R$

$Maximum momentum, p_{\max} = R$

$Total energy = E_{2}$

$E_{2} = \frac{1}{2} m ω_{2}^{2} R^{2}$ ...(iii)

$p_{\max} = m v_{\max} = m ω_{2} R$

$R = m ω_{2} R \Rightarrow m ω_{2} = 1$ ...(iv)

From eqns. (ii) and (iv),

$\frac{a}{b} = \frac{ω_{2}}{ω_{1}}$ ...(v)

From eqns. (i) and (iii),

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

If $\frac{a}{b} = n^{2}$ and $\frac{a}{R} = n$ then from eqn. (v),

$\frac{ω_{2}}{ω_{1}} = n^{2}$

and from eqn. (vi),

$\frac{E_{1}}{E_{2}} = \frac{ω_{1}^{2}}{ω_{2}^{2}} \times n^{2} = \frac{ω_{1}}{ω_{2}}$

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

Solution

1 $E_{1} ω_{1} = E_{2} ω_{2}$

2 $\frac{ω_{2}}{ω_{1}} = n^{2}$

3 $ω_{1} ω_{2} = n^{2}$

4 $\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

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Solution

1 E1ω1=E2ω2

2 ω2ω1=n2

3 ω1ω2=n2

4 E1ω1=E2ω2

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1 $E_{1} ω_{1} = E_{2} ω_{2}$

2 $\frac{ω_{2}}{ω_{1}} = n^{2}$

3 $ω_{1} ω_{2} = n^{2}$

4 $\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$