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Current Question (ID: 15947)

Question:

$\text{The mathematical forms for three sinusoidal traveling waves are given by:}$ $\text{Wave 1: } y(x, t) = (2 \text{ cm}) \sin(3x - 6t)$ $\text{Wave 2: } y(x, t) = (3 \text{ cm}) \sin(4x - 12t)$ $\text{Wave 3: } y(x, t) = (4 \text{ cm}) \sin(5x - 11t)$ $\text{where } x \text{ is in meters and } t \text{ is in seconds. Of these waves:}$

Options:

1. $\text{Wave 1 has the highest wave speed as well as the maximum transverse string speed.}$
2. $\text{Wave 2 has the highest wave speed, while Wave 1 has the maximum transverse string speed.}$
3. $\text{Wave 3 has the highest wave speed as well as the maximum transverse string speed.}$
4. $\text{Wave 2 has the highest wave speed, while Wave 3 has the maximum transverse string speed.}$

Solution:

$\text{Hint: } v_{p\text{max}} = A\omega$ $\text{Step: Find the maximum and minimum speed of the wave.}$ $\text{The maximum particle speed of the wave is given by: } v_{p\text{max}} = A\omega$ $\text{The maximum particle speed of the wave 1 is given by:}$ $v_{p\text{max}} = 2 \times 6 = 12 \text{ m/s}$ $\text{The maximum particle speed of the wave 2 is given by:}$ $v_{p\text{max}} = 3 \times 12 = 36 \text{ m/s}$ $\text{The maximum particle speed of the wave 3 is given by:}$ $v_{p\text{max}} = 4 \times 11 = 44 \text{ m/s}$ $\text{The wave speed of the wave is given by: } v = \frac{\omega}{k}$ $\text{The wave speed of the wave 1 is given by: } v_1 = \frac{6}{3} = 2 \text{ m/s}$ $\text{The wave speed of the wave 2 is given by: } v_2 = \frac{12}{4} = 3 \text{ m/s}$ $\text{The wave speed of the wave 3 is given by: } v_3 = \frac{11}{5} = 2.2 \text{ m/s}$ $\text{Therefore, the wave 2 has the highest wave speed, while the wave 3 has the maximum transverse string speed.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}