Question

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 figures. If $\frac{a}{b}=n^2$ and $\frac{a}{R}=n$, then the correct equation(s) is (are)

• A. $E_1{\omega }_1=E_2{\omega }_2$
• B.
  $\frac{{\omega }_2}{{\omega }_1}=n^2$
• C.
  ${\omega }_1{\omega }_2=n^2$
• D.
  $\frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$

Answer 1

Answer: (B), (D)

$\frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$
First oscillator:
maximum momentum will be attained in the first graph when $x=0$
Momentum at $b$ is = $n\times a\times {\omega }_1$
$\frac{b}{a}=n\times {\omega }_{!}$

In secound case,
Momentum $\Rightarrow R=n\times R\times {\omega }_{2}n{\omega }_2=1$
$\frac{{\omega }_2}{{\omega }_1}=\frac{a}{b}=n^2$
Hence, option B is correct.

calculation of energies:
for case first; $E_1=\frac{1}{2}{m}_1{a}^2{\omega }_1^2$
for case secound; $E_2=\frac{1}{2}{m}_2{R}^2{\omega }_2^2$
$\frac{E_1}{E_2}=\frac{\frac{1}{2}{m}_1{a}^2{\omega }_1^2}{\frac{1}{2}{m}_2{R}^2{\omega }_2^2}$
where, $\frac{a}{R}=n$ and $\frac{{\omega }_2}{{\omega }_1}=n^2$
$\Rightarrow \frac{E_1}{E_2}=n^2\times \frac{1}{(n^2{)}^2}\frac{E_1}{E_2}=\frac{1}{n^2}$
where, $\frac{1}{n^2}=\frac{{\omega }_1}{{\omega }_2}$
So, $\frac{E_1}{E_2}=\frac{{\omega }_1}{{\omega }_2}$

Hence, option D is correct.