JEE Advance - Physics (2011 - Paper 1 Offline - No. 18)
The phase space diagram for simple harmonic motion is a circle centred at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then
Explanation
Energy of simple harmonic oscillator is
$$E = {1 \over 2}k{A^2}$$
where k is the force constant and A the amplitude of the oscillator. Since the oscillator is the same, the value of k is the same. Hence
$${E_1} = {1 \over 2}kA_1^2$$ and $${E_2} = {1 \over 2}kA_2^2$$
$$\therefore$$ $${{{E_1}} \over {{E_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^2}$$
Now, A1 = maximum value of displacement of oscillator having energy E1 = 2a and A2 = a. Therefore
$${{{E_1}} \over {{E_2}}} = {\left( {{{2a} \over a}} \right)^2} = 4$$. So, $${E_1} = 4{E_2}$$
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