WBJEE 2024 Math Paper

The particle will come to rest when its velocity is zero. To find the velocity, we differentiate the position function with respect to time:$v = \frac{dx}{dt} = a \sqrt{\lambda} \cos (\sqrt{\lambda} t + b)$Setting the velocity to zero, we get:$a \sqrt{\lambda} \cos (\sqrt{\lambda} t + b) = 0$This equation is satisfied when:$\cos (\sqrt{\lambda} t + b) = 0$The cosine function is zero at odd multiples of $\; \frac{\pi}{2}$. Therefore, we have: $\sqrt{\lambda} t + b = (2n+1) \; \frac{\pi}{2}$, where $n$ is an integer.Solving for $t$, we get:$t = \frac{(2n+1)\frac{\pi}{2} - b}{\sqrt{\lambda}}$ This gives us an infinite number of times when the particle comes to rest. Let's consider the first two instances ( $n = 0$ and $n = 1$ ):$t_1 = \frac{\frac{\pi}{2} - b}{\sqrt{\lambda}}$$t_2 = \frac{\frac{3\pi}{2} - b}{\sqrt{\lambda}}$ Now, we can find the positions of the particle at these times by plugging them back into the original position function:$x_1 = a \sin (\sqrt{\lambda} t_1 + b) = a \sin \left(\frac{\pi}{2}\right) = a$$x_2 = a \sin (\sqrt{\lambda} t_2 + b) = a \sin \left(\frac{3\pi}{2}\right) = -a$The distance between these two points is:$\left| x_2 - x_1 \right| = \left| -a - a \right| = 2a$Therefore, the correct answer is Option C: 2a.