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Current Question (ID: 19977)

Question:

$\text{Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by}$ $y = 10 \cos(\pi x) \sin\left(\frac{2\pi t}{T}\right) \text{ cm. The amplitude of the particle at }$ $x = \frac{4}{3} \text{ cm will be:}$

Options:

1. $5 \text{ cm}$
2. $10 \text{ cm}$
3. $15 \text{ cm}$
4. $20 \text{ cm}$

Solution:

$\text{Hint: } y = A \cos(kx) \sin(\omega t)$ $\text{Step: Find the amplitude of the particle at } x = \frac{4}{3} \text{ cm.}$ $\text{The equation of a standing wave is given by;} y = A \cos(kx) \sin(\omega t).$ $\text{Where } A \text{ is the amplitude of the wave, } k \text{ is the wave number, and } \omega \text{ is the angular frequency.}$ $\text{From the given equation, } y = 10 \cos(\pi x) \sin\left(\frac{2\pi t}{T}\right) \text{ cm.}$ $\text{Now compare to the standard standing wave equation } y = A \cos(kx) \sin(\omega t), \text{ we get;}$ $\text{The amplitude } A = 10 \text{ cm, wave number } k = \pi, \text{ and the angular frequency } \omega = \frac{2\pi}{T}. $ $\text{Substitute the value of } x, \text{ the amplitude of the particle at } x = \frac{4}{3} \text{ cm.}$ $\text{The amplitude of the particle at a specific position } x \text{ in a stationary wave is given by the term } A \cos(kx).$ $\text{Amplitude}$ $= A \cos(kx) \Rightarrow 10 \cdot \cos\left(\frac{4\pi}{3}\right) = 10 \cdot \left(-\frac{1}{2}\right) = -5 \text{ cm.}$ $\text{Since amplitude is a measure of displacement and is always taken as a positive quantity.}$ $\text{Therefore, the amplitude of the particle at } x = \frac{4}{3} \text{ cm is } 5 \text{ cm.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}