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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

The position co-ordinates of a particle moving in a \(3-D\) coordinates system is given by \(x = a\,cos\,\omega t\) , \(y = a\,sin\,\omega t\) and \(z = a\omega t\) The speed of the  particle is

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Answer & Solution

Correct Answer

(A) \(\sqrt 2 \,a\omega\)

Step-by-step Solution

Detailed explanation

\(\mathrm{v}_{\mathrm{x}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=-\mathrm{a} \omega \sin \omega \mathrm{t}\)

\(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a\omega} \cos \omega \mathrm{t}\)

\(\mathrm{v}_{\mathrm{z}}=\frac{\mathrm{d} z}{\mathrm{dt}}=\mathrm{a} \omega\)

\(\therefore \quad \mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}}=\mathrm{a} \omega \sqrt{2}\)

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