A body of mass ' m ' performs linear S.H.M. given by equation x = Psin ω t + Qsin (ω t + π/2). The total energy of the particle at any instant is (a) 1/2mω² PQ (b) 1/2(mω²)/(P²Q²) (c) 1/2mω²(P²+Q²) (d) 1/2mω²P²Q²

Answer: (c). x = Psin ω t + Qcos ω t: two perpendicular-phase SHMs of the same ω. Resultant amplitude A = √(P² + Q²) E = 1/2mω²A² = 1/2mω²(P² + Q²) So the answer is (c).

A body of mass ' m ' performs linear S.H.M. given by equation x =… | MHT CET 2021 (22 Sep, Shift 2)

Q50MHT CET 2021 · 22 SEP · SHIFT 2Easy

Energy in SHM

A body of mass ' $m$ ' performs linear S.H.M. given by equation $x = P\sin\omega t + Q\sin\left(\omega t + \frac{\pi}{2}\right)$ . The total energy of the particle at any instant is

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Answer: (c)

$x = P\sin\omega t + Q\cos\omega t$ : two perpendicular-phase SHMs of the same $\omega$ .
Resultant amplitude $A = \sqrt{P^2 + Q^2}$
$E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{1}{2}m\omega^2(P^2 + Q^2)$
So the answer is (c).

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Energy in SHM question from MHT CET 2021 (22 Sep, Shift 2)

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