The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
Option: 1 \mathrm{\pi \ rad}
Option: 2 \mathrm{\frac{3\pi}{2} \ rad}
Option: 3 \mathrm{\frac{\pi}{2} \ rad}
Option: 4 zero

Answers (1)

$$\begin{aligned} &\text { SHM is represented by }\ displacement= \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}\\ &\text { Then, the velocity and acceleration after differentiation will be }\\ &\mathbf{v}=\mathbf{A} \omega \cos \omega \mathbf{t}\\ &\mathrm{a}=-\mathrm{A} \omega^{2} \sin \omega \mathrm{t}=\mathrm{A} \omega^{2} \sin (\omega \mathrm{t}+\pi) \end{aligned}

So the phase difference between x & a= \mathrm{\pi \ rad}

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