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प्रश्न
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
विकल्प
simple harmonic
on a straight line
on a circle
with constant acceleration
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उत्तर
on a circle
We know,
\[\frac{\text {d}^2}{\text {dt}^2} \overrightarrow{r} = - \omega^2 \overrightarrow{r} \]
But there is a phase difference of 90o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.
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