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 the 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)

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

Maximum linear momentum in case $1$ is $(p_1{)}_{\text{max}}=m{v}_{\text{max}}$
From given figure 1,
$v_{max}=a{\omega }_1$ and $(p_1{)}_{\text{max}}=b$
$\therefore b=ma{\omega }_1$ ........ (1)
Similarly, maximum linear momentum in case $2$ is $(p_2{)}_{\text{max}}=m{v}_{\text{max}}$

From given figure 2,
$v_{max}=R{\omega }_1$ and $(p_2{)}_{\text{max}}=R$
$\therefore R=mR{\omega }_2$
$\Rightarrow 1=m{\omega }_2$ ..........(2)

Dividing (1) by (2),
$\frac{b}{1}=\frac{a{\omega }_1}{{\omega }_2}$
$\therefore \frac{{\omega }_2}{{\omega }_1}=\frac{a}{b}=n^2$ ............(3)
[given $\frac{a}{b}=n^2$]
$\therefore$ $\left(b\right)$ is a correct option.

Also $E_1=\frac{1}{2}m{\omega }_1^2{a}^2$
and $E_2=\frac{1}{2}m{\omega }_2^2{R}^2$
$\therefore \frac{E_1}{E_2}=\frac{{\omega }_1^2}{{\omega }_2^2}\times \frac{a^2}{R^2}=\frac{{\omega }_1^2}{{\omega }_2^2}\times n^2$
$\Rightarrow \frac{E_1}{E_2}=\frac{{\omega }_1}{{\omega }_2}\times n^2\times \frac{1}{n^2}$ [from Eq.(3)]
$\Rightarrow \frac{E_1}{E_2}=\frac{{\omega }_1}{{\omega }_2}$
$\therefore \frac{E_1}{{\omega }_1}=\frac{E_2}{{\omega }_2}$
$\left(d\right)$ is also a correct option.