The linear displacement 'x' of the bob of simple pendulum from its mean position varies as…

The linear displacement 'x' of the bob of simple pendulum from its mean position varies as $\mathrm{x}=\mathrm{a} \sin \left(\frac{\pi}{\sqrt{2}} \mathrm{t}\right)$ where 'a' is its amplitude expressed in metre and 't' is in second. The length of simple pendulum is (Take ' $\mathrm{g}^{\prime}=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}$ )
  1. $1 \cdot 5 \mathrm{~m}$
  2. $3 \cdot 0 \mathrm{~m}$
  3. $2 \cdot 0 \mathrm{~m}$
  4. $2 \cdot 5 \mathrm{~m}$

Solution

$\mathrm{x}=\mathrm{a} \sin \left(\frac{\pi}{\sqrt{2}} \mathrm{t}\right)$ $\therefore \omega=\frac{\pi}{\sqrt{2}}=\frac{2 \pi}{\mathrm{T}} \quad \therefore \mathrm{T}=2 \sqrt{2}$ $\mathrm{~T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$ $2 \sqrt{2}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$ $2=\pi^{2} \frac{\ell}{\mathrm{g}}$ $\therefore \quad \ell=\frac{2 \mathrm{~g}}{\pi^{2}}=2 \mathrm{~m}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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