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Current Question (ID: 19952)

Question:

$\text{The displacement of a damped harmonic oscillator is given by}$ $x(t) = e^{-0.1t} \cos(10\pi t + \phi).$ $\text{Here } t \text{ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:}$

Options:

1. $4 \text{ s}$
2. $13 \text{ s}$
3. $7 \text{ s}$
4. $27 \text{ s}$

Solution:

$\text{The amplitude of the damped oscillator is given by } A(t) = e^{-0.1t}.$ $\text{To find the time when the amplitude drops to half, we set } \frac{1}{2} = e^{-0.1t}.$ $\text{Taking natural logarithm on both sides, we get } \ln\left(\frac{1}{2}\right) = -0.1t.$ $t = \frac{\ln(2)}{0.1} \approx 6.93 \text{ s}.$ $\text{Therefore, the closest option is } 7 \text{ s}.$

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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
}