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

Correct answer: (C) 2a. Hint : dx/dt=a cos (√(λ) t+b) · √(λ)=0 [ at rest, dx/dt=0 ] ⇒ cos (√(λ) t+b)=0 ⇒ √(λ) t+b=(π)/2, (-π)/2 ∴ x_1 =a sin (π)/2 and x_2=a sin ((-π)/2 ) =a =-a ∴ Distance between the points = |x_2-x_1 |=2 a

Q60 / 155 · WBJEE 2024 · Mathematics

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]

A
a
B
$\frac{a}{2}$
Correct answer
C
2a
D
4a

Solution

Hint : $\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \cos (\sqrt{\lambda} \mathrm{t}+\mathrm{b}) \cdot \sqrt{\lambda}=0\left[\begin{array}{l}\text { at rest, } \\ \frac{\mathrm{dx}}{\mathrm{dt}}=0\end{array}\right] \Rightarrow \cos (\sqrt{\lambda} \mathrm{t}+\mathrm{b})=0$ $\begin{aligned} \Rightarrow & \sqrt{\lambda} \mathrm{t}+\mathrm{b}=\frac{\pi}{2}, \frac{-\pi}{2} \\ \therefore \quad \mathrm{x}_1 & =\mathrm{a} \sin \frac{\pi}{2} \quad \text { and } \mathrm{x}_2=\mathrm{a} \sin \left(\frac{-\pi}{2}\right) \\ & =\mathrm{a} \quad=-\mathrm{a} \end{aligned}$ $\therefore \quad$ Distance between the points $=\left|\mathrm{x}_2-\mathrm{x}_1\right|=2 \mathrm{a}$

Hence, the correct answer is option: 'C'.

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]

Solution

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