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If vectors \mathop \text{A}\limits^ \to = \cos \omega \text{t}\,\hat i + \sin \omega \text{t}\,\hat j and \mathop \text{B}\limits^ \to = \cos \frac{{\omega \text{t}}} {\text{2}}\,\hat i + \sin \frac{{\omega \text{t}}} {\text{2}}\,\hat j are functions or time, then the value of t at which they are orthogonal to each other is :

  • Option 1)

    t = \frac{\pi } {{2\omega }}

  • Option 2)

    t = \frac{\pi } {\omega }

  • Option 3)

    t = 0

  • Option 4)

    t = \frac{\pi } {{4\omega }}

 

Answers (1)

best_answer

As we learnt in

Composition of two SHM in perpendicular direction -

x=A_{1}sin omega t

y=A_{2}sin left ( omega t+delta ight )

 

 

- wherein

Resultant equation

frac{x^{2}}{A{_{1}}^{2}}+frac{y{_{2}}^{2}}{A{_{2}}^{2}}-frac{2xycos delta }{A_{1}A_{2}}= sin ^{2}delta

 

 \vec{A}.\vec{B}= (\cos \omega t \hat i + \sin \omega t \hat{j}). \left ( \cos \frac{\omega t}{2}\:\hat{i}+ \sin \frac{\omega t}{2} \hat{j} \right )=\cos \left ( \omega t- \frac{\omega t}{2}\right )

When they are orthogonal, \vec{A}.\vec{B}=0

\cos \left( \omega t - \frac {\omega t}{2} \right )=0

\cos \left( \frac {\omega t}{2} \right )=0 \:\:\:\:\:\:\:\:\: \Rightarrow \frac{\omega t}{2}=\frac{\pi}{2}

t=\frac{\pi}{\omega}

Correct option is 2.

 

 

 


Option 1)

t = \frac{\pi } {{2\omega }}

This option is incorrect 

Option 2)

t = \frac{\pi } {\omega }

This option is correct 

Option 3)

t = 0

This option is incorrect 

Option 4)

t = \frac{\pi } {{4\omega }}

This option is incorrect 

Posted by

Aadil

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