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Current Question (ID: 16016)

Question:

$\text{If the transverse displacement of a string clamped at both ends is given by } y(x, t) = (12 \text{ cm}) \sin(6.28x) \cos(3.14t), \text{ where } x \text{ is in cm and } t \text{ is in seconds, then which of the following is not true?}$

Options:

1. $\text{The velocity of the component wave is } 0.5 \text{ cm/s.}$
2. $\text{The amplitude of one of the component waves is } 6 \text{ cm.}$
3. $\text{The distance between two consecutive nodes is } 0.5 \text{ cm.}$
4. $x = 0.25 \text{ cm is the first node except the nodes at the ends.}$

Solution:

$\text{Hint: In a string clamped at both ends, standing waves are formed.}$ $\text{Step 1: Write the equation of standing waves.}$ $y = 2A \sin(kx) \cos \omega t$ $\text{Step 2: Compare the equations.}$ $\text{By comparing the general equation of transverse wave and the equation given in the question, we get;}$ $\omega = 3.14 \text{ rad/s}$ $k = 6.28 \text{ cm}^{-1} = \frac{2\pi}{\lambda}$ $2A = 12 \text{ cm}$ $\text{Step 3: Find the desired quantities.}$ $v = \frac{\omega}{k} = 0.5 \text{ cm/s}$ $A = 6 \text{ cm}$ $\frac{\lambda}{2} = 0.5 \text{ cm}$ $y = 2A \sin kx \cos \omega t$ $\sin kx = 0$ $kx = n\pi$ $2\pi x = n\pi$ $x = \frac{n}{2}$ $x = 0.5 \text{ cm}$ $\text{Therefore, option (4) is not True.}$ $\text{Hence, option (4) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}