The displacement of a particle executing simple harmonic motion is given by y = A0 + A sin ω t + B cos ω t. Then the amplitude of its oscillation is given by: (a) √(A² + B²) (b) A + B (c) A + √(A² + B²) (d) √(A₀² + (A + B)²)

Answer: (a). A₀ only shifts the mean position; it does not change the amplitude. Asin ω t + Bcos ω t = √(A² + B²) sin (ω t + δ) So the answer is (a) √(A² + B²).

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Periodic Motion & SHM Basics

The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sin $\omega$ t + B cos $\omega$ t.

Then the amplitude of its oscillation is given by:

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Answer: (a)

$A_0$ only shifts the mean position; it does not change the amplitude.
$A\sin\omega t + B\cos\omega t = \sqrt{A^2 + B^2}\,\sin(\omega t + \delta)$
So the answer is (a) $\sqrt{A^2 + B^2}$ .

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