Import Question JSON

Current Question (ID: 19895)

Question:

$\text{A particle executes simple harmonic motion with a time period } T. \text{ It starts at its equilibrium position at } t = 0. \text{ How will the graph of its kinetic energy } (KE) \text{ versus time } (t) \text{ look like?}$

Options:

1. $\begin{array}{c} KE \\ 0 \\ T \\ t \end{array}$
2. $\begin{array}{c} KE \\ 0 \\ \frac{T}{2} \quad T \\ t \end{array}$
3. $\begin{array}{c} KE \\ 0 \\ \frac{T}{2} \quad T \\ t \end{array}$
4. $\begin{array}{c} KE \\ 0 \\ \frac{T}{4} \quad \frac{T}{2} \quad T \\ t \end{array}$

Solution:

$\text{Hint: } KE = \frac{1}{2} m A^2 \omega^2 \cos^2 \omega t$ $\text{Step: Find the graph of the kinetic energy versus time for the particle.}$ $\text{The displacement of the particle is given by:}$ $x = A \sin \omega t$ $\text{The velocity of the particle performing the SHM is given by:}$ $v = \frac{dx}{dt} = A \omega \cos \omega t$ $\text{The kinetic energy of the particle at any time } t \text{ is given by:}$ $K.E = \frac{1}{2} m A^2 \omega^2 \cos^2 \omega t$ $\Rightarrow K.E = \frac{1}{2} m A^2 \omega^2 \cos^2 \frac{2 \pi}{T} t$ $\text{Therefore, the graph of kinetic energy versus time for the particle forms a squared cosine function, as shown in the figure below:}$ $\begin{array}{c} KE \\ 0 \\ \frac{T}{4} \quad \frac{T}{2} \quad T \\ t \end{array}$ $\text{Hence, option (4) is the correct answer.}$

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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
}