Animated Solution for Physics - Kinematics: Two vectors $\mathbf{A}$ and $\mathbf{B}$ are defined as $\mathbf{A}=a\hat{\mathbf{i}}$ and $\mathbf{B}=a(\cos \omega t\hat{\mathbf{i}}+\sin \omega t\hat{\mathbf{j}})$, where $a$ is a constant and $\omega =\pi /6{\text{ rad s}}^{-1}$. If $|\mathbf{A}+\mathbf{B}|=\sqrt{3}|\mathbf{A}-\mathbf{B}|$ at time $t=\tau$ for the first time, the value of $\tau$, in seconds, is .............

Imagine you are standing at the origin of a coordinate system. You have two arrows in your hands. One arrow, vector $\mathbf{A}$, is firmly planted along the x-axis, never changing its length or direction. The other arrow, vector $\mathbf{B}$, is identical in length but is sweeping around you in a circle, like the second hand of a clock, with a constant angular velocity $\omega$. This is the beautiful dynamic setup of our problem.

When we add or subtract two vectors, we are geometrically finding the diagonals of the parallelogram they form. Since both $\mathbf{A}$ and $\mathbf{B}$ have the same magnitude $a$, they actually form a rhombus!

Let's write down the math. We are given: $\mathbf{A}=a\hat{\mathbf{i}}$ $\mathbf{B}=a\cos (\omega t)\hat{\mathbf{i}}+a\sin (\omega t)\hat{\mathbf{j}}$

The sum and difference of these vectors are: $\mathbf{A}+\mathbf{B}=(a+a\cos \omega t)\hat{\mathbf{i}}+a\sin \omega t\hat{\mathbf{j}}$ $\mathbf{A}-\mathbf{B}=(a-a\cos \omega t)\hat{\mathbf{i}}-a\sin \omega t\hat{\mathbf{j}}$

To find their magnitudes, we square the components and add them. Squaring is a brilliant mathematical move here because it clears out the square roots and sets the stage for some trigonometric elegance.

$|\mathbf{A}+\mathbf{B}{|}^2=a^2(1+\cos \omega t{)}^2+a^2{\sin }^2\omega t$ $|\mathbf{A}-\mathbf{B}{|}^2=a^2(1-\cos \omega t{)}^2+a^2(-\sin \omega t{)}^2$

I know expanding these brackets might look tedious, but let's take a breath and watch the magic happen. When we expand $(1\pm \cos \omega t{)}^2$, we get $1\pm 2\cos \omega t+{\cos }^2\omega t$.

Notice that we also have a ${\sin }^2\omega t$ term waiting outside. By the fundamental identity of trigonometry, ${\cos }^2\omega t+{\sin }^2\omega t=1$.

This collapses our massive equations into something beautifully simple: $|\mathbf{A}+\mathbf{B}{|}^2=2a^2(1+\cos \omega t)$ $|\mathbf{A}-\mathbf{B}{|}^2=2a^2(1-\cos \omega t)$

Now, we deploy a favorite tool of physicists and mathematicians alike: the half-angle formulas. We know that $1+\cos \theta =2{\cos }^2(\theta /2)$ and $1-\cos \theta =2{\sin }^2(\theta /2)$. Applying these, we get perfect squares!

$|\mathbf{A}+\mathbf{B}{|}^2=4a^2{\cos }^2\left(\frac{\omega t}{2}\right)$ $|\mathbf{A}-\mathbf{B}{|}^2=4a^2{\sin }^2\left(\frac{\omega t}{2}\right)$

The problem states a very specific condition: the length of one diagonal is exactly $\sqrt{3}$ times the length of the other. $|\mathbf{A}+\mathbf{B}|=\sqrt{3}|\mathbf{A}-\mathbf{B}|$

Squaring both sides to use our simplified expressions: $|\mathbf{A}+\mathbf{B}{|}^2=3|\mathbf{A}-\mathbf{B}{|}^2$

Substitute the half-angle expressions: $4a^2{\cos }^2\left(\frac{\omega t}{2}\right)=3\times 4a^2{\sin }^2\left(\frac{\omega t}{2}\right)$

Cancel out the $4a^2$ and rearrange to find the tangent: ${\tan }^2\left(\frac{\omega t}{2}\right)=\frac{1}{3}$ $\tan \left(\frac{\omega t}{2}\right)=\pm \frac{1}{\sqrt{3}}$

This is a standard trigonometric equation. The general solution is: $\frac{\omega t}{2}=n\pi \pm \frac{\pi }{6}$

We are given $\omega =\pi /6$. Let's plug that in: $\frac{\pi }{12}t=n\pi \pm \frac{\pi }{6}$ $t=12n\pm 2$

We are looking for the first time this happens, which means we need the smallest positive value for $t$. Setting $n=0$ and taking the positive sign gives us exactly $t=2$ seconds.

Geometrically, at $t=2$ seconds, the angle between the vectors is $60^{\circ }$. The diagonals of a rhombus with a $60^{\circ }$ angle have lengths in the exact ratio of $\sqrt{3}:1$. The math perfectly mirrors the physical reality!