JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \(\sqrt 2 \,a\omega\)
- B \(a\omega \)
- C \(\sqrt 3 \,a\omega\)
- D \(2\,a\omega \)
Answer & Solution
Correct Answer
(A) \(\sqrt 2 \,a\omega\)
Step-by-step Solution
Detailed explanation
\(\mathrm{v}_{\mathrm{x}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=-\mathrm{a} \omega \sin \omega \mathrm{t}\) \(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a\omega} \cos \omega \mathrm{t}\)…
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