Question 8, Review Exercise

Solutions of Question 8 of Review Exercise of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

When particle is at a position of $\sqrt{2}+i \sqrt{2}$nm from its mean position calculate its amplitude when $\theta=45^{\circ}$.

Solution.

Here we have $$x= \sqrt{2} + i \sqrt{2}, \quad \theta=\dfrac{\pi}{4}.$$ We have to find $x_{\max}$. By using the formula $$\begin{align} &x=x_{\max} e^{i\theta} \\ \implies & \sqrt{2} + i \sqrt{2}=x_{\max} e^{i\dfrac{\pi}{4}} \\ \implies & x_{\max} \left(\cos\dfrac{\pi}{4} +i\sin\dfrac{\pi}{4}\right)=\sqrt{2} + i \sqrt{2} \\ \implies & x_{\max} \left(\dfrac{1}{\sqrt{2}} +i\dfrac{1}{\sqrt{2}}\right)=\sqrt{2} + i \sqrt{2} \\ \implies & \dfrac{x_{\max}}{2} \left(\sqrt{2} +i\sqrt{2}\right)=\sqrt{2} + i \sqrt{2}= \\ \implies & x_{\max}=2. \end{align}$$ Therefore, the amplitude is $2$nm.

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