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Current Question (ID: 19908)

Question:

$\text{The equation: } Y = A \sin(\omega t + \phi_0) \text{ represents the time-displacement relation of simple harmonic motion (SHM). At } t = 0,$ $\text{the displacement of the particle is } Y = \frac{A}{2}, \text{ and it is moving in the negative } x\text{-direction. The initial phase angle } \phi_0 \text{ is:}$

Options:

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

Solution:

$\text{Hint: } y = A \sin(\omega t + \phi_0), \ v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi_0).$ $\text{Step: Find the initial phase angle } \phi_0.$ $\text{The equation of the SHM is given by;}$ $y = A \sin(\omega t + \phi_0)$ $\text{At } t = 0, \text{ the displacement is } y = \frac{A}{2}, \text{ and the particle is moving along the negative } x\text{-direction then we get;}$ $\frac{A}{2} = A \sin(\phi_0)$ $\Rightarrow \sin(\phi_0) = \frac{1}{2}$ $\Rightarrow \phi_0 = \sin^{-1}\left(\frac{1}{2}\right)$ $\Rightarrow \phi_0 = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$ $\text{The velocity of the particle is given by;}$ $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi_0)$ $\text{At } t = 0, \text{ the velocity of the particle is given by;}$ $v = A\omega \cos(\phi_0)$ $\text{As the particle is moving along the negative } x\text{-direction, so } v < 0.$ $\cos(\phi_0) < 0$ $\text{Therefore, the initial phase angle is } \frac{5\pi}{6}.$ $\text{Hence, option (3) is the correct answer.}$

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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
}