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Current Question (ID: 11237)

Question:
$\text{Equation of a simple harmonic motion is given by } x = a \sin \omega t. \text{ For which value of } x, \text{ kinetic energy is equal to the potential energy?}$
Options:
  • 1. $x = \pm a$
  • 2. $x = \pm \frac{a}{2}$
  • 3. $x = \pm \frac{a}{\sqrt{2}}$ (Correct)
  • 4. $x = \pm \frac{\sqrt{3}a}{2}$
Solution:
$\text{Hint: Kinetic energy is maximum at the mean position whereas potential energy is maximum at the extreme position.}$ $\text{Step 1: Write the kinetic and potential energy of the particle performing SHM.}$ $K = \frac{1}{2}m\omega^2\left(A^2 - x^2\right)$ $U = \frac{1}{2}kx^2$ $\text{Step 2: Find the value of } x, \text{ when kinetic energy is equal to the potential energy.}$ $K = U$ $\frac{1}{2}m\omega^2\left(A^2 - x^2\right) = \frac{1}{2}kx^2$ $\Rightarrow \omega = \sqrt{\frac{k}{m}}$ $\Rightarrow m\omega^2 = k$ $\Rightarrow A^2 - x^2 = x^2$ $\Rightarrow x = \pm \frac{A}{\sqrt{2}}$ $\text{Hence option (3) is the correct answer.}$

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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
}