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Current Question (ID: 19892)

Question:

$\text{A simple harmonic oscillator of angular frequency } 2 \text{ rad s}^{-1} \text{ is acted upon by an external force } F = \sin t \text{ N.}$ $\text{If the oscillator is at rest in its equilibrium at } t = 0, \text{ its position at later times is proportional to:}$

Options:

1. $\cos t - \frac{1}{2} \sin 2t$
2. $\sin t + \frac{1}{2} \cos 2t$
3. $\sin t + \frac{1}{2} \sin 2t$
4. $\sin t - \frac{1}{2} \sin 2t$

Solution:

$\text{Hint: } a = \frac{dv}{dt}$ $\text{Step 1: Find the restoring force.}$ $\text{The equation of position of SHO is given by;} x = A \sin(\omega t)$ $\text{Now, find the equation of velocity of SHO;} v = \frac{dx}{dt} = A \omega \cos(\omega t)$ $\text{Now, find the equation of acceleration of SHO;} a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \quad \cdots (1)$ $\text{Now, apply Newton's 2nd equation of motion;} F_{\text{res}} = ma = -m A \omega^2 \sin \omega t$ $\text{By using equation (1) we get the restoring force as;} F_{\text{res}} = -m A \omega^2 \sin \omega t$ $\text{Step 2: Find the net force acting on SHO.}$ $F_{\text{net}} = F_{\text{res}} + F_{\text{ext}}$ $\text{Therefore;} F_{\text{net}} = -m A \omega^2 \sin \omega t + \sin(t) \quad \cdots (2)$ $\text{Step 3: Find the net acceleration of SHO.}$ $\text{Again apply Newton's 2nd law of motion;} a_{\text{net}} = \frac{F_{\text{net}}}{m}$ $\text{Now, by using equation (2);} a_{\text{net}} = \frac{-m A \omega^2 \sin \omega t + \sin(t)}{m}$ $\text{Simplify.}$ $a_{\text{net}} = -A \omega^2 \sin(\omega t) + \frac{1}{m} \sin(t) \quad \cdots (3)$ $\text{Step 4: Find the position function at any time } t.$ $\text{Integrate equation (3) with respect to time;} v' = \int A \omega \cos(\omega t) - \frac{1}{m} \cos(t) + C_1 \quad \cdots (3)$ $\text{By using given initial condition, at } t = 0; \Rightarrow v = 0. \text{Hence;} 0 = A \omega - \frac{1}{m} \cos(0) + C_1 = A \omega - \frac{1}{m}$ $\text{Therefore, equation (3) becomes;} v' = A \sin(\omega t) - \frac{1}{m} \sin(t) + A \omega - \frac{t}{m} + C_2 \quad \cdots (4)$ $\text{By using given initial condition, at } t = 0; \Rightarrow x = 0. \text{Hence;} 0 = 0 - 0 + 0 + 0 + C_2 \Rightarrow C_2 = 0$ $\text{Therefore, equation (3) becomes;} x' = A \sin(\omega t) - \frac{1}{m} \sin(t) + A \omega - \frac{t}{m}$ $\text{Now, substitute the known values;} x' = A \sin(2t) - \frac{1}{m} \sin(t) + 2At - \frac{t}{m}$ $\text{By taking the negative sign common;} x' = \left(-\frac{1}{m} \sin(t) - A \sin(2t)\right) + 2At - \frac{t}{m}$ $\text{Therefore;} x' \propto \left(\sin(t) - \frac{1}{2} \sin(2t)\right)$ $\text{Hence, option (4) is the correct answer.}$

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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
}