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Current Question (ID: 15948)

Question:

$\text{A wave in a string has an amplitude of } 2 \text{ cm. The wave travels in the positive direction of the } x\text{-axis with a speed of } 128 \text{ m/s and it is noted that } 5 \text{ complete waves fit in the } 4 \text{ m length of the string. The equation describing the wave is:}$

Options:

1. $y = (0.02 \text{ m}) \sin(7.85x + 1005t)$
2. $y = (0.02 \text{ m}) \sin(15.7x - 2010t)$
3. $y = (0.02 \text{ m}) \sin(15.7x + 2010t)$
4. $y = (0.02 \text{ m}) \sin(7.85x - 1005t)$

Solution:

$\text{Hint: } y = a \sin(\omega t - kx)$ $\text{Step 1: Find the angular frequency of the wave.}$ $\text{It is given that the amplitude of the wave in a string is, i.e. } A = 2 \text{ cm}$ $\text{And also, the wave is travelling in the positive x-direction}$ $\text{And the velocity, } v = 128 \text{ m/s}$ $\text{According to the question, 5 complete waves fit in a 4 m length of the string}$ $\text{which means that:}$ $5\lambda = 4 \ldots (i)$ $\text{From (i):}$ $\lambda = 4/5$ $\text{And the propagation constant, } k = \frac{2\pi}{\lambda} = 2 \times \frac{5\pi}{4} = 7.85$ $\text{Also, the phase velocity of the wave is given by:}$ $v = \frac{\omega}{k} = 128 \text{ ms}^{-1}$ $\Rightarrow \omega = v \times k = 128 \times 7.85$ $= 1004.8 \text{ which can be approx. written as } = 1005 \text{ rad/sec.}$ $\text{Step 2: Fine the equation of the wave.}$ $\text{The equation of wave can be represented as:}$ $y = A\sin(kx - \omega t)$ $\text{By putting the given values in the above equation, we get:}$ $y = 2\sin(7.85x - 1005t)$ $= (0.02)\sin(7.85x-1005t) \quad [\text{amplitude in meter } = 0.02 \text{ m}]$ $\text{Or, } y = (0.02)\text{m } \sin (7.85x-1005t)$ $\text{Hence, option (4) is the correct answer.}$

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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
}