Ratio of Total Mechanical Energies from Phase Space Diagram of SHM

For a simple harmonic oscillator, the total mechanical energy $E$ is related to the amplitude of oscillation $A$ by the relation:
$E = \frac{1}{2} k A^2$
where $k$ is the force constant of the oscillator.

From the phase space diagram (Momentum vs. Position):

• The maximum displacement (amplitude) for the inner trajectory corresponding to energy $E_2$ is $A_2 = a$.
• The maximum displacement (amplitude) for the outer trajectory corresponding to energy $E_1$ is $A_1 = 2a$.

Since both trajectories represent the same oscillator, the spring constant $k$ is identical in both cases.

Thus, the total mechanical energies are given by:
$E_2 = \frac{1}{2} k a^2$
$E_1 = \frac{1}{2} k (2a)^2 = 4 \left(\frac{1}{2} k a^2\right) = 4 E_2$

Therefore, $E_1 = 4 E_2$.

Correct Option: (C)