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Simple Harmonic Motion question

2011 · Shift 1 · Q63
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Simple Harmonic Motion question

2011 · Shift 1 · Q63

JEE AdvancedPhysicsSimple Harmonic MotionMCQ+3 / −1
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) 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 int he figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English ComprehensionThe phase space diagram for a ball thrown vertically up from ground is
  1. A
    IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English Option 1
  2. B
    IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English Option 2
  3. C
    IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English Option 3
  4. D
    IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 17 English Option 4
View written solutionFree

Correct answer: D

  1. Choose coordinates in phase space

For one-dimensional motion, phase space has:

  • horizontal axis: position xxx
  • vertical axis: momentum ppp

For a ball thrown vertically upward from the ground:

  • take upward as positive,
  • ground as x=0x=0x=0,
  • initial momentum is positive.

So initially the ball starts at

\qquad p=p_0>0.$$ 2. **Write momentum as a function of position** The ball moves under constant gravitational force downward: $$F=-mg.$$ Using energy conservation, $$\frac{p^2}{2m}+mgx=\frac{p_0^2}{2m}.$$ Hence, $$p^2=p_0^2-2m^2gx.$$ So, $$p=\pm \sqrt{p_0^2-2m^2gx}.$$ This is the equation of the phase curve. 3. **Interpret the shape** From $$p^2=p_0^2-2m^2gx,$$ we get $$x=\frac{p_0^2-p^2}{2m^2g}.$$ This is a **left-opening parabola** in the $x$-$p$ plane. Important points: - At launch: $$(x,p)=(0,p_0)$$ - At highest point: $$p=0,\quad x=\frac{p_0^2}{2m^2g}$$ - On coming down to ground: $$(x,p)=(0,-p_0)$$ Thus the trajectory goes from the point $(0,p_0)$ downward-right to the turning point $(x_{\max},0)$ and then downward-left to $(0,-p_0)$. 4. **Direction of time flow** As time increases: - $x$ first increases and $p$ decreases from $+p_0$ to $0$, - then $x$ decreases while $p$ becomes negative. So the arrow should go from the upper point on the $p$-axis to the rightmost point, then back to the lower point on the $p$-axis. 5. **Match with the options** Therefore the correct phase-space diagram must be: - a **left-opening parabola**, - starting at $(0,+p_0)$, - passing through the rightmost turning point where $p=0$, - ending at $(0,-p_0)$, - with arrows showing this direction. This corresponds to **Option D**. 6. **Comparison with stored answer** Stored correct answer: **D** My derived answer: **D** They agree.
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