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A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:

1. πa/T
2. 3π²a/T
3. πa√3/T
4. πa√3/2T

Correct answer: πa√3/T

Solution

The speed of a simple harmonic oscillator at displacement x is given by v = ω√(A² - x²), where ω = 2π/T. Substituting A = a, x = a/2, and ω = 2π/T, we get v = (2π/T)√(a² - (a/2)²) = πa√3/T.

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