Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension. For such systems, 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) versus 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 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 a ball thrown vertically up from ground is :

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)`