Kinematics 6 Question 9
9. Two vectors $A$ and $B$ are defined as $A=a \hat{i}$ and $B=a(\cos \omega t \hat{i}+\sin \omega t \hat{j})$, where $a$ is a constant and $\omega=\pi / 6$ $\operatorname{rad~s}^{-1}$. If $|A+B|=\sqrt{3}|A-B|$ at time $t=\tau$ for the first time, the value of $\tau$ in seconds, is
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Answer:
Correct Answer: 9. (2.0)
Solution:
- $A=a \hat{i}$ and $B=a \cos \omega \hat{i}+a \sin \omega \hat{j}$
$$ \begin{aligned} & A+B=(a+a \cos \omega t) \hat{i}+a \sin \omega t \hat{j} \\ & A-B=(a-a \cos \omega t) \hat{i}+a \sin \omega t \hat{j} \\ & |A+B|=\sqrt{3}|A-B| \\ & \sqrt{(a+a \cos \omega t)^{2}+(a \sin \omega t)^{2}}=\sqrt{3} \\ & \qquad \sqrt{(a-a \cos \omega t)^{2}+(a \sin \omega t)^{2}} \\ & \Rightarrow \quad 2 \cos \frac{\omega t}{2}= \pm \sqrt{3} \times 2 \sin \frac{\omega t}{2} \Rightarrow \tan \frac{\omega t}{2}= \pm \frac{1}{\sqrt{3}} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad \frac{\omega t}{2}=n \pi \pm \frac{\pi}{6} \quad \Rightarrow \quad \frac{\pi}{12} t=n \pi \pm \frac{\pi}{6} \\ & \Rightarrow \quad t=(12 n \pm 2) s=2 s, 10 s, 14 s \text { and so on. } \end{aligned} $$
