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Current Question (ID: 11252)

Question:

$\text{Two simple harmonic motions, } y_1 = a \sin \omega t \text{ and } y_2 = 2a \sin \left(\omega t + \frac{2\pi}{3}\right) \text{ are superimposed on a particle of mass } m\text{. The maximum kinetic energy of the particle will be:}$

Options:

1. $\frac{1}{2}m\omega^2 a^2$
2. $\frac{5}{4}m\omega^2 a^2$
3. $\frac{3}{2}m\omega^2 a^2$ (Correct)
4. $\text{Zero}$

Solution:

$\text{Hint: The maximum kinetic energy of the particle is given as } K_{\text{max}} = \frac{1}{2}m\omega^2 A^2\text{.}$ $\text{Step 1: Find the resultant amplitude of the vectors.}$ $A = \sqrt{a^2 + (2a)^2 + 2(a)(2a)\cos \frac{2\pi}{3}}$ $A = \sqrt{3a}$ $\text{Step 2: Calculate the maximum kinetic energy of the particle.}$ $K_{\text{max}} = \frac{1}{2}m\omega^2 A^2 = \frac{3m\omega^2 a^2}{2}\text{.}$ $\text{Hence, option (3) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}