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Current Question (ID: 19933)

Question:

$\text{For a particle performing linear SHM, its position } (x) \text{ as a function of time } (t) \text{ is given by } x = A \sin(\omega t + \delta).$ $\text{If, at } t = 0, \text{ particle is at } +\frac{A}{2} \text{ and is moving towards } x = +A, \text{ then } \delta = $

Options:

1. $\frac{\pi}{3} \text{ rad}$
2. $\frac{\pi}{6} \text{ rad}$
3. $\frac{\pi}{4} \text{ rad}$
4. $\frac{5\pi}{6} \text{ rad}$

Solution:

$\text{Hint: Use a phasor diagram.}$ $\text{Step 1: Draw the phasor diagram for the given SHM.}$ $\text{Step 2: Find the phase of the SHM.}$ $\text{In the phasor diagram}$ $\sin \delta = \frac{\frac{A}{2}}{A} = \frac{1}{2}$ $\delta = \frac{\pi}{6} \text{ radian}$ $\text{Hence, option (2) is the correct answer.}$

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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
}