Phase space diagrams are useful tools in analysing all kond of dynamical problems. Theya re especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical system in one-dimelnsion. for such systeam, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is `x(t)` vs. `p(t)` curve in this plane. The arrow on the curve indicates the time flow. for example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum. upwards (or to right) is positive and downwards (or to left) is negative.

The phase space diagram for simple harmonic motion is a circle cen-tered at the origin.In the figure, the two circles represent the same oscillator but for different initial conditions, and `E_(1)` for `E_(2)` are the total mechanical energies respectively. Then

Options

A. `E_(1) = sqrt(2)E_(2)`
B. `E_(1) = 2E_(2)`
C. `E_(1) = 4 E_(2)`
D. `E_(1) = 16 E_(2)`

Answer

C

Explanation

In `1st` case amplitude of `SHM` is `a`.
In `2nd` case amplitude of `SHM` is `2a`
Total energy `= (1)/(2)K(amplitude)^(2)`
`E_(1)/(2)2k(2a)^(2) , E_(1) = 4 E_(2)`
Alternative :
Linear momentum `P = mv`
`= momegasqrt(A^(2) - x^(2))`
`rArr P^(2) = m^(2)omega^(2) (A^(2) - x^(2))`
`rArr P^(2) + m^(2)omega^(2)x^(2) = m^(2)omega^(2)A^(2)`
Equation of circle (smaller)
`P^(2) + x^(2) = a^(2) .......(iii)`
Comparing `(i)` and `(ii)`
Amplitude `A = 2a`
and `(momega)^(2) = (1)/(m)`
`(1)/(2)momega^(2)(A)^(2)`
So energy `E_(1) = (1)/(2) momega^(2) (2a)^(2)`
`(1)/(2)(1)/(m)xx(4a^(2))`
`= (2a^(2))/(m)`
Comparing `(i)` and `(iii)`
`(momega)^(2) = 1 rArr momega^(2)A^(2) = (1)/(2) xx (1)/(m)a^(2) = (1)/(2)(a^(2))/(m^(2))(1a^(2))/(2 m)`. So, `(E_(1))/(E_(2)) = 4 rArr E_(1) = 4E_(2)`