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The resultant of two rectangular simple harmonic motion of the same frequency and unequal amplitudes but differing in phase by \frac{\pi}{}2 is

  • Option 1)

    Simple harmonic

  • Option 2)

    Circular

  • Option 3)

    Elliptical

  • Option 4)

    Parabolic

 

Answers (1)

best_answer

If the first equation is y_{1} = a_{1}\sin\omega t \Rightarrow \sin\omega t = \frac{y_{1}}{a_{1}}

Then second equation will be y_{2} = a_{2}\sin(\omega t + \frac{\pi}{2})\Rightarrow \cos\omega t = \frac{y_{2}}{a_{2}}

By squaring and adding both equations :

\sin^{2}\omega t + \cos^{2}\omega t = \frac{y_{1}^{2}}{a_{1}^{2}} + \frac{y_{2}^{2}}{a_{2}^{2}} \\*\Rightarrow \frac{y_{1}^{2}}{a_{1}^{2}} + \frac{y_{2}^{2}}{a_{2}^{2}} = 1

This equation is of ellipse.


Option 1)

Simple harmonic

This is incorrect.

Option 2)

Circular

This is incorrect.

Option 3)

Elliptical

This is correct.

Option 4)

Parabolic

This is incorrect.

Posted by

Aadil

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