For a particle performing linear S.H.M. of amplitude 'r', the potential energy is ' λ ' times its total energy. The displacement of particle is (a) rλ (b) (r)/(λ) (c) r√(λ) (d) (r)/(√(λ))

Answer: (c). PE/E = (1/2kx²)/(1/2kr²) = (x²)/(r²) = λ x = r√(λ) So the answer is (c).

For a particle performing linear S.H.M. of amplitude 'r', the… | MHT CET 2026 (17 Apr, Shift 2)

Q4MHT CET 2026 · 17 APR · SHIFT 2Easy

Energy in SHM

For a particle performing linear S.H.M. of amplitude 'r', the potential energy is ' $\lambda$ ' times its total energy. The displacement of particle is

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Answer: (c)

$\dfrac{PE}{E} = \dfrac{\frac{1}{2}kx^2}{\frac{1}{2}kr^2} = \dfrac{x^2}{r^2} = \lambda$
$x = r\sqrt{\lambda}$
So the answer is (c).

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