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Current Question (ID: 11245)

Question:

$\text{A particle of mass } m \text{ oscillates in simple harmonic motion between points } X_1 \text{ and } X_2, \text{ the equilibrium position being } O. \text{ Its kinetic energy is correctly represented by which of the following graphs?}$

Options:

1. $\text{Graph 1: Shows a parabola opening downward with maximum at equilibrium position } O \text{ and minimum at extreme positions } X_1 \text{ and } X_2$ (Correct)
2. $\text{Graph 2: Shows a sinusoidal wave with minimum at equilibrium position } O \text{ and maximum at extreme positions } X_1 \text{ and } X_2$
3. $\text{Graph 3: Shows a straight line passing through all three points } X_1, O, \text{ and } X_2$
4. $\text{Graph 4: Shows a parabola opening upward with minimum at equilibrium position } O \text{ and maximum at extreme positions } X_1 \text{ and } X_2$

Solution:

$\text{Hint: Kinetic energy is maximum at the mean position.}$ $\text{Step: Find the correct plot for the kinetic energy.}$ $\text{The kinetic energy for a particle of mass } m \text{ oscillating in simple harmonic motion is given by; } K.E = \frac{1}{2}m\omega^2(A^2 - x^2)$ $\text{At mean position } (x = 0) \text{ at point } O, \text{ is given by; } K.E = \frac{1}{2}m\omega^2 A^2.$ $\text{Therefore, the graph (1) represents the correct variation of kinetic energy at the mean position.}$ $\text{Hence, option (1) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}