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Current Question (ID: 11240)

Question:

$\text{The kinetic energy (K) of a simple harmonic oscillator varies with displacement (x) as shown. The period of the oscillation will be: (mass of oscillator is 1 kg)}$

Options:

1. $\frac{\pi}{2} \text{ s}$ (Correct)
2. $\frac{1}{2} \text{ s}$
3. $\pi \text{ s}$
4. $1 \text{ s}$

Solution:

$\text{Hint: At the mean position, particle will have maximum kinetic energy.}$ $\text{Step 1: Find the angular frequency of the particle.}$ $(P.E.)_{max} = (K.E.)_{max} = 2$ $\Rightarrow \frac{1}{2}m\omega^2A^2 = 2$ $\Rightarrow \omega^2 = \frac{4}{(0.5)^2} = 16$ $\Rightarrow \omega = 4 \text{ rad/sec}$ $\text{Step 2: Find the time period of the particle in SHM.}$ $\text{Time period} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ sec}.$ $\text{Hence, option (1) is the correct answer.}$

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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
}