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Current Question (ID: 11297)

Question:

$\text{The figure shows the circular motion of a particle. The radius of the circle, the period, the sense of revolution, and the initial position are indicated in the figure. The simple harmonic motion of the } x \text{-projection of the radius vector of the rotating particle } P \text{ will be:}$

Options:

1. $x(t) = B\text{sin}\left(\frac{2\pi t}{30}\right)$ (Correct)
2. $x(t) = B\text{cos}\left(\frac{\pi t}{15}\right)$
3. $x(t) = B\text{sin}\left(\frac{\pi t}{15} + \frac{\pi}{2}\right)$
4. $x(t) = B\text{cos}\left(\frac{\pi t}{15} + \frac{\pi}{2}\right)$

Solution:

$\text{Hint: Use the trigonometric functions to find the equation. Step 1: Find the horizontal projection of the radius vector. Let the angular velocity of the particle executing circular motion be } \omega \text{ and when it is at } Q \text{, it makes an angle } \theta \text{ as shown in the diagram. Now, we can write } OR = OQ\text{cos}(90 - \theta) \Rightarrow OR = OQ\text{sin}\theta = OQ\text{sin}\omega t \text{ } [OQ=r, \theta=\omega t] \Rightarrow x = r\text{sin}\omega t \text{ Step 2: Find the equation of motion. The amplitude is } B \text{ so } x = B\text{sin}\omega t \text{. The angular frequency } (\omega) \text{ is related to the period by: } \omega = \frac{2\pi}{T} = \frac{2\pi}{30}\text{ rad/s} \Rightarrow x = B\text{sin}\left(\frac{2\pi t}{30}\right) \text{. Hence, option (1) is the correct answer.}$

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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
}