Discussion Forum

 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 $E_{1}$, 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 equations is/are

Answer:

Explanation:

 Ist particle

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

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

$\Rightarrow$   $mv_{max}=b\Rightarrow v_{max}=\frac{b}{m}$

  $E_{1}=\frac{1}{2}mv^{2}_{max}=\frac{m}{2}\left[\frac{b}{m}\right]^{2}=\frac{b^{2}}{2m}$

$A_{1}\omega_{1}=v_{max}=\frac{b}{m}$

$\Rightarrow$      $\omega_{1}=\frac{b}{ma}=\frac{1}{mn^{2}}\left(A_{1}=a, \frac{a}{b}=n^{2}\right)$

  II nd particle

  P=0 at x= R  $\Rightarrow$ A₂ = R

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

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

   $A_{2}\omega_{2}=\frac{R}{m}\Rightarrow \omega_{2}=\frac{R}{mR}=\frac{1}{m}$

  (b)  

                $\frac{\omega_{2}}{\omega_{1}}=\frac{1/m}{1/mn^{2}}=n^{2}$

(c)           $\omega_{1}{\omega_{2}}=\frac{1}{mn^{2}}\times\frac{1}{m}=\frac{1}{m^{2}n^{2}}$

  (d)  $\frac{E_{1}}{\omega_{1}}=\frac{b^{2/}2m}{1/mn^{2}}=\frac{b^{2}n^{2}}{2}=\frac{a^{2}}{2n^{2}}=\frac{R^{2}}{2}$

             $\frac{E_{2}}{\omega_{2}}=\frac{R^{2}/2m}{1/m}=\frac{R^{2}}{2}$

   $\Rightarrow$   $\frac{E_{1}}{\omega_{1}}=\frac{E_{2}}{\omega_{2}}$

 Note:    It is not given that the second figure is a circle . But from the figure and as per the requirement of question , we consider it is a circle