MHT CET · Physics · Oscillations
The path length of oscillation of simple pendulum of length \(1 \mathrm{~m}\) is \(16 \mathrm{~cm}\). its maximum velocity is (Take \(g=\pi^2 \mathrm{~m} / \mathrm{s}^2\) )
- A \(2 \pi \mathrm{cm} / \mathrm{s}\)
- B \(8 \pi \mathrm{cm} / \mathrm{s}\)
- C \(4 \pi \mathrm{cm} / \mathrm{s}\)
- D \(16 \pi \mathrm{cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(8 \pi \mathrm{cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Amplitude of SHM, \(a=8 \mathrm{~cm}\)
Maximum velocity is given by, \(v=a \omega=a \sqrt{\frac{g}{l}}=8 \times \sqrt{\frac{\pi^2}{1}}=8 \pi \mathrm{cm} / \mathrm{s}\)
Maximum velocity is given by, \(v=a \omega=a \sqrt{\frac{g}{l}}=8 \times \sqrt{\frac{\pi^2}{1}}=8 \pi \mathrm{cm} / \mathrm{s}\)
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