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(C) The velocity of the particle is given by the derivative of position with respect to time: $v = \frac{dx}{dt} = a \cos (\sqrt{\lambda} t + b) \cdot

Vedclass · Accepted Answer

$2a$

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(C) The velocity of the particle is given by the derivative of position with respect to time: $v = \frac{dx}{dt} = a \cos (\sqrt{\lambda} t + b) \cdot \sqrt{\lambda}$. For the particle to come to rest, the velocity must be zero: $v = 0 \Rightarrow a \sqrt{\lambda} \cos (\sqrt{\lambda} t + b) = 0$. Since $a 
eq 0$ and $\lambda > 0$, we have $\cos (\sqrt{\lambda} t + b) = 0$. This occurs when the argument is $\pm \frac{\pi}{2}$. Let the two points be $x_1$ and $x_2$: $x_1 = a \sin (\frac{\pi}{2}) = a(1) = a$. $x_2 = a \sin (-\frac{\pi}{2}) = a(-1) = -a$. The distance between these two points is $|x_1 - x_2| = |a - (-a)| = |2a| = 2a$.

If a particle moves in a straight line according to the law $x = a \sin (\sqrt{\lambda} t + b)$, then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

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