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
Pick an option to check your answer.
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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