JEE Advance - Physics (2015 - Paper 1 Offline - No. 18)

Two independent harmonic oscillators of equal masses are oscillating about the origin with angular frequencies $$\omega$$1 and $$\omega$$2 and have total energies E1 and E2, respectively. The variations of their momenta p with positions x are shown in the figures. If $${a \over b} = {n^2}$$ and $${a \over R} = n$$, then the correct equations is/are
JEE Advanced 2015 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English
E1$$\omega$$1 = E2$$\omega$$2
$${{{\omega _2}} \over {{\omega _1}}} = {n^2}$$
$${\omega _1}{\omega _2} = {n^2}$$
$${{{E_1}} \over {{\omega _1}}} = {{{E_2}} \over {{\omega _2}}}$$

Explanation

1st Particle

P = 0 at x = a $$\Rightarrow$$ 'a' is the amplitude of oscillation 'A1'.

At x = 0, P = b (at mean position)

$$ \Rightarrow m{v_{\max }} = b \Rightarrow {v_{\max }} = {b \over m}$$

$${E_1} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{b \over m}} \right]^2} = {{{b^2}} \over {2m}}$$

$${A_1}{\omega _1} = {v_{\max }} = {b \over m}$$

$$ \Rightarrow {\omega _1} = {b \over {ma}} = {1 \over {m{n^2}}}(A = a,\,{a \over b} = {n^2})$$

2nd Particle

P = 0 at x = R $$\Rightarrow$$ A2 = R

At x = 0, P = R $$\Rightarrow$$ $${v_{\max }} = {R \over m}$$

$${E_2} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{R \over m}} \right]^2} = {{{R^2}} \over {2m}}$$

$${A_2}{\omega _2} = {R \over m} \Rightarrow {\omega _2} = {R \over {mR}} = {1 \over m}$$

(b) $${{{\omega _2}} \over {{\omega _1}}} = {{1/m} \over {1/m{n^2}}} = {n^2}$$

(c) $${\omega _1}{\omega _2} = {1 \over {m{n^2}}} \times {1 \over m} = {1 \over {{m^2}{n^2}}}$$

(d) $${{{E_1}} \over {{\omega _1}}} = {{{b^2}/2m} \over {1/m{n^2}}} = {{{b^2}{n^2}} \over 2} = {{{a^2}} \over {2{n^2}}} = {{{R^2}} \over 2}$$

$${{{E_2}} \over {{\omega _2}}} = {{{R^2}/2m} \over {1/m}} = {{{R^2}} \over 2}$$

$$ \Rightarrow {{{E_1}} \over {{\omega _1}}} = {{{E_2}} \over {{\omega _2}}}$$

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