The path length of oscillation of simple pendulum of length $1 \mathrm{~m}$ is $16 \mathrm{~cm}$. its…

The path length of oscillation of simple pendulum of length $1 \mathrm{~m}$ is $16 \mathrm{~cm}$. its maximum velocity is (Take $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$ )
  1. $2 \pi \mathrm{cm} / \mathrm{s}$
  2. $8 \pi \mathrm{cm} / \mathrm{s}$
  3. $4 \pi \mathrm{cm} / \mathrm{s}$
  4. $16 \pi \mathrm{cm} / \mathrm{s}$

Solution

Amplitude of SHM, $a=8 \mathrm{~cm}$ Maximum velocity is given by, $v=a \omega=a \sqrt{\frac{g}{l}}=8 \times \sqrt{\frac{\pi^2}{1}}=8 \pi \mathrm{cm} / \mathrm{s}$ *

Asked in: MHT CET 2022 (11 Aug Shift 1)

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