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Question

A particle is subjected to two mutually perpendicular simple harmonic motions such that its $$x$$ and $$y$$ coordinates are given by $$x=2\sin{\omega t}$$ ; $$y=2\sin {\left(\omega t + \dfrac{\pi}{4}\right)}$$. The path of the particle will be:


A
an ellipse
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B
a straight line
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C
a parabola
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D
a circle
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Solution

The correct option is A an ellipse
$$x=2\sin { \omega t } ;y=2\sin { \left( \omega t+\frac { \pi  }{ 4 }  \right)  }$$
$$\Rightarrow \sin { \omega t } =\dfrac { x }{ 2 } $$     .....$$(I)$$   
$$\Rightarrow \cos { \omega t } =\sqrt { 1-\sin ^{ 2 }{ \omega t }  } =\sqrt { 1-{ \left( \dfrac { x }{ 2 }  \right)  }^{ 2 } } =\sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 }  } $$     .....$$(II)$$
$$y=2\sin { \left( \omega t+\dfrac { \pi  }{ 4 }  \right)  } \\ \Rightarrow y=2\dfrac { \left( \sin { \omega t } +\cos { \omega t }  \right)  }{ \sqrt { 2 }  } =\sqrt { 2 } \left( \sin { \omega t } +\cos { \omega t }  \right) $$
using the values of from $$(I)$$ and $$(II)$$, we have
$$y=\sqrt { 2 } \left( \dfrac { x }{ 2 } +\sqrt { 1-\frac { { x }^{ 2 } }{ 4 }  }  \right) \\ \Rightarrow \dfrac { y }{ \sqrt { 2 }  } -\dfrac { x }{ 2 } =\sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 }  } \\ \\ \Rightarrow \dfrac { { y }^{ 2 } }{ 2 } +\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { { xy } }{ \sqrt { 2 }  } =1-\dfrac { { x }^{ 2 } }{ 4 } \\ \\ \Rightarrow \dfrac { { x }^{ 2 } }{ 2 } +\dfrac { { y }^{ 2 } }{ 2 } -\dfrac { { xy } }{ \sqrt { 2 }  } -1=0\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\sqrt { 2 } xy-2=0$$
which is an equation of an oblique ellipse.

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Physics

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