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Current Question (ID: 19894)

Question:
$\text{A particle performs simple harmonic motion with amplitude } A. \text{ Its speed is tripled at the instant that it is at a distance } \frac{2A}{3} \text{ from the equilibrium position. The new amplitude of the motion is:}$
Options:
  • 1. $\frac{A}{3}\sqrt{41}$
  • 2. $3A$
  • 3. $A\sqrt{3}$
  • 4. $\frac{7A}{3}$
Solution:
$\text{Hint: } v = \omega \sqrt{A^2 - x^2}$ $\text{Velocity at any position is given by}$ $v = \omega \sqrt{A^2 - x^2}$ $\text{at}$ $x = \frac{2A}{3}$ $v = \omega \sqrt{A^2 - \frac{4A^2}{9}} = v = \frac{\sqrt{5}}{3} A \omega$ $\text{if speed at } x = \frac{2A}{3} \text{ is tripled i.e.}$ $v' = 3v = \sqrt{5} A \omega$ $\text{So new amplitude } A' \text{ is}$ $v' = \omega \sqrt{A'^2 - \left(\frac{4A}{3}\right)^2}$ $\sqrt{5} A \omega = \omega \sqrt{A'^2 - \frac{4A^2}{9}}$ $A' = \frac{7A}{3}$

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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
}