Single Choice

A particle executing SHM. The phase difference between acceleration and displacement is:

A$$0$$
B$$\dfrac{\pi}{2}$$
C$$ \pi$$
Correct Answer
D$$\dfrac{3}{2} \pi$$

Solution

Let the displacement of a particle executing simple harmonic motion at any instant t is, $$x = A cos \omega t$$ Velocity, $$v = \dfrac{dx}{dt} = \dfrac{d}{dt}(A cos \omega t) = -A \omega sin \omega t$$ Acceleration, $$a = \dfrac{dv}{dt} .= -A \omega^2 cos \omega t = A \omega^2 cos(\omega t + \pi)$$ Phase of displacement, $$\phi_1 = \omega t$$ Phase of acceleration, $$\phi_2 = \omega t + \pi$$ $$\therefore$$ Phase difference = $$\phi_2 - \phi_1 = (\omega t + \pi) - \omega t = \pi$$

SIMILAR QUESTIONS

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