Question 376: Oscillations

Question

Two independent harmonic oscillators of equal masses 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 ${a \over b} = {n^2}$ and ${a \over R} = n$, then the correct equations is/are
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l11z64zo/a88cca7d-3a6b-408a-99b1-cc129e548840/28377840-a9c8-11ec-8265-8513bb2a18eb/file-1l11z64zp.png?format=png)

Accepted Answer

1st Particle
P = 0 at x = a $\Rightarrow$ 'a' is the amplitude of oscillation 'A
1
'.
At x = 0, P = b (at mean position)
$ \Rightarrow m{v_{\max }} = b \Rightarrow {v_{\max }} = {b \over m}$
${E_1} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{b \over m}} ight]^2} = {{{b^2}} \over {2m}}$
${A_1}{\omega _1} = {v_{\max }} = {b \over m}$
$ \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 $\Rightarrow$ A
2
= R
At x = 0, P = R $\Rightarrow$ ${v_{\max }} = {R \over m}$
${E_2} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{R \over m}} ight]^2} = {{{R^2}} \over {2m}}$
${A_2}{\omega _2} = {R \over m} \Rightarrow {\omega _2} = {R \over {mR}} = {1 \over m}$
(b) ${{{\omega _2}} \over {{\omega _1}}} = {{1/m} \over {1/m{n^2}}} = {n^2}$
(c) ${\omega _1}{\omega _2} = {1 \over {m{n^2}}} 	imes {1 \over m} = {1 \over {{m^2}{n^2}}}$
(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}}}$

Answer

E
1
$\omega$
1
= E
2
$\omega$
2

Answer

${{{\omega _2}} \over {{\omega _1}}} = {n^2}$

Answer

${\omega _1}{\omega _2} = {n^2}$

Answer

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

Oscillations

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