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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Physics
Chapter
Oscillations

JEE Main 2015 · 10 April · Online · Physics · Oscillations

xandy displacements of a particle are given as x(t) = asin ω...

xandyxandy displacements of a particle are given as x(t)=asinωtandy(t)=asin2\omegat.x\left(t\right) = asin \omega tandy\left(t\right) = asin 2\omegat . Its trajectory will look like:xandy displacements of a particle are given as x(t) = asin ω tandy(t) = asin 2 . Its trajectory will look like:xandy displacements of a particle are given as x(t) = asin ω tandy(t) = asin 2 . Its trajectory will look like:xandy displacements of a particle are given as x(t) = asin ω tandy(t) = asin 2 . Its trajectory will look like:xandy displacements of a particle are given as x(t) = asin ω tandy(t) = asin 2 . Its trajectory will look like:

Options

  1. AThe incorrect statement regarding conformations of ethane is:
  2. BThe incorrect statement regarding conformations of ethane is:
  3. CThe incorrect statement regarding conformations of ethane is:
  4. DThe incorrect statement regarding conformations of ethane is:

Correct answer

C. The incorrect statement regarding conformations of ethane is:

Step-by-step solution

∵x=Asin\omegat⇒sin\omegat=xA∵ x = Asin \omegat ⇒ sin\omegat = \dfrac{x}{A}

Also, cos\omegat=1−sin⁡2\omegat=1−x2A2cos\omegat = \sqrt{1 - \sin^{2} \omegat} = \sqrt{1 - \dfrac{{x}^{2}}{{A}^{2}}}

⇒cos\omegat=A2−x2A⇒ cos \omegat = \dfrac{\sqrt{{A}^{2} - {x}^{2}}}{A}

As, y=2Asinωtcos\omegaty = 2Asin \omega tcos \omegat

⇒y=2AxAA2−x2A⇒ y = 2A\dfrac{x}{A}\dfrac{\sqrt{{A}^{2} - {x}^{2}}}{A}

⇒y=2AxA2−x2⇒ y = \dfrac{2}{A}x\sqrt{{A}^{2} - {x}^{2}}

⇒y=0atx=0andx=±A⇒ y = 0atx = 0andx = \pm A

Which is possible only in option (C)
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