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JEE Mains · Physics · STD 11 - 3.2 motion in plane

\(x-y\) समतल में कण की गति निम्न समीकरणो \(x =4 \sin \left(\frac{\pi}{2}-\omega t \right) m\) तथा \(y =4 \sin (\omega t ) m\). द्वारा व्यक्त की जाती है। कण का पथ होगा-

  1. A वृत्ताकार
  2. B हेलीकल
  3. C परवलयाकार
  4. D दीर्घवृत्ताकार
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Answer & Solution

Correct Answer

(A) वृत्ताकार

Step-by-step Solution

Detailed explanation

\(x=4 \sin \left(\frac{\pi}{2}-\omega t\right) y=4 \cos (\omega t)\) \(x=4 \cos (\omega t) \quad y=4 \sin (\omega t)\) Eliminate ' \(t\) ' to find relation between \(x\) and \(y\) \(x^{2}+y^{2}=y^{2} \cos ^{2} \omega t+y^{2} \sin ^{2} \omega t=4^{2}\) \(x^{2}+y^{2}=4^{2}\)
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