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Simple Harmonic Motion question

2011 · Shift 1 · Q64
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Simple Harmonic Motion question

2011 · Shift 1 · Q64

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 15 English ComprehensionThe 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 IIT-JEE 2011 Paper 1 Offline Physics - Simple Harmonic Motion Question 15 English
  1. A
    E1 = 2\sqrt22​ E2
  2. B
    E1 = 2E2
  3. C
    E1 = 4E2
  4. D
    E1 = 16E2
View written solutionFree

Correct answer: C

  1. Phase space equation for SHM

For a simple harmonic oscillator, x=Acos⁡(ωt+ϕ),p=mx˙=−mωAsin⁡(ωt+ϕ).x=A\cos(\omega t+\phi), \qquad p=m\dot x=-m\omega A\sin(\omega t+\phi).x=Acos(ωt+ϕ),p=mx˙=−mωAsin(ωt+ϕ).

Eliminating time, x2A2+p2(mωA)2=1.\frac{x^2}{A^2}+\frac{p^2}{(m\omega A)^2}=1.A2x2​+(mωA)2p2​=1.

So in the usual xxx-ppp phase plane, the trajectory is an ellipse. If the axes are scaled suitably, it appears as a circle. In either case, the size of the curve is proportional to amplitude AAA.

  1. Energy of SHM

Total mechanical energy is E=12kA2=12mω2A2.E=\frac12 kA^2=\frac12 m\omega^2 A^2.E=21​kA2=21​mω2A2.

Hence, E∝A2.E\propto A^2.E∝A2.

Also, in phase space, the intercepts are proportional to AAA and mωAm\omega AmωA, so if one phase curve has radius/intercept twice that of the other, its energy becomes four times.

  1. Using the figure

From the given phase-space diagram, the larger circle has radius twice the smaller one. Therefore, A1A2=2.\frac{A_1}{A_2}=2.A2​A1​​=2.

Thus, E1E2=(A1A2)2=22=4.\frac{E_1}{E_2}=\left(\frac{A_1}{A_2}\right)^2=2^2=4.E2​E1​​=(A2​A1​​)2=22=4.

So, E1=4E2.E_1=4E_2.E1​=4E2​.

  1. Check options
  • A: E1=2E2E_1=\sqrt2 E_2E1​=2​E2​ ❌
  • B: E1=2E2E_1=2E_2E1​=2E2​ ❌
  • C: E1=4E2E_1=4E_2E1​=4E2​ ✅
  • D: E1=16E2E_1=16E_2E1​=16E2​ ❌

Therefore, the correct option is C.

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