A simple pendulum performs simple harmonic motion about x = 0 with an amplitude 'A' and angular frequency ω. The speed of pendulum at x = A/8 will be. (a) (√(61)Aω)/(8) (b) (√(62)Aω)/(8) (c) (√(63)Aω)/(8) (d) (√(65)Aω)/(8)

Answer: (c). v = ω√(A² - x²) v = ω√(A² - A²/64) = ω A√(63/64) So the answer is (c) (√(63)Aω)/(8).

A simple pendulum performs simple harmonic motion about x = 0 with an… | MHT CET 2026 (20 May, Shift 1)

Q1MHT CET 2026 · 20 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude 'A' and angular frequency $\omega$ . The speed of pendulum at $x = \frac{A}{8}$ will be.

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Answer: (c)

$v = \omega\sqrt{A^2 - x^2}$
$v = \omega\sqrt{A^2 - \frac{A^2}{64}} = \omega A\sqrt{\frac{63}{64}}$
So the answer is (c) $\frac{\sqrt{63}A\omega}{8}$ .

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