The bob of simple pendulum of length ' $L$ ' is released from a position of small angular displacement…

The bob of simple pendulum of length ' $L$ ' is released from a position of small angular displacement $\theta$. Its linear displacement at time ' $\mathrm{t}$ ' is ( $\mathrm{g}=$ acceleration due to gravity)
  1. $L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$
  2. $\mathrm{L} \theta \sin \left[2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \cdot \mathrm{t}\right]$
  3. $\mathrm{L} \theta \cos \left[2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \cdot \mathrm{t}\right]$
  4. $L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]$

Solution

Equation for displacement for a particle performing S.H.M. is, $\mathrm{y}=\mathrm{A} \cos \omega \mathrm{t}=\mathrm{L} \cos \left(\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}\right)$ But time period of a simple pendulum, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\therefore \quad y=L \cos \left(\frac{2 \pi}{2 \pi \sqrt{\frac{L}{g}} \times t}\right)=L \cos \left(\sqrt{\frac{g}{L}} \times t\right)$ $\therefore \quad$ The linear displacement is, $\mathrm{s}=\mathrm{y} \theta=L \theta \cos \left[\sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}\right]$

Asked in: MHT CET 2023 (14 May Shift 1)

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