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
B
a straight line
C
a parabola
D
a circle

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.Physics

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