Question

If a particle moves in a straight line according to the law \(x = a \sin(\sqrt{t} + b)\), then the particle will come to rest at two points whose distance is:

• A. \(a\)
• B. \(\frac{a}{2}\)
• C. \(2a\)
• D. \(4a\)

Hint:

When dealing with trigonometric functions in particle motion problems, set the function equal to zero to find the points of rest.

Answer 1

The Correct Option is C

The particle’s position is given by:

\[ x = a \sin(\sqrt{t} + b). \]

Step 1: Velocity of the particle. The velocity \( v \) is the derivative of \( x \) with respect to time \( t \):

\[ v = \frac{dx}{dt}. \]

Differentiate \( x \):

\[ v = a \cos(\sqrt{t} + b) \cdot \frac{d}{dt}(\sqrt{t} + b). \]

Since \( \frac{d}{dt}(\sqrt{t} + b) = \frac{1}{2\sqrt{t}} \), the velocity becomes:

\[ v = a \cos(\sqrt{t} + b) \cdot \frac{1}{2\sqrt{t}}. \]

Step 2: Condition for the particle to come to rest. The particle comes to rest when \( v = 0 \), i.e., when:

\[ \cos(\sqrt{t} + b) = 0. \]

The general solution for \( \cos(\theta) = 0 \) is:

\[ \sqrt{t} + b = \frac{\pi}{2} + n\pi \quad \text{for integers } n. \]

From this, we solve for \( t \):

\[ \sqrt{t} = \frac{\pi}{2} + n\pi - b. \]

Let the particle come to rest at two consecutive points corresponding to \( n = k \) and \( n = k + 1 \). The values of \( \sqrt{t} \) at these points are:

\[ \sqrt{t_1} = \frac{\pi}{2} + k\pi - b, \quad \sqrt{t_2} = \frac{\pi}{2} + (k + 1)\pi - b. \]

Step 3: Distance between rest points. At the rest points, the position \( x \) is:

\[ x_1 = a \sin(\sqrt{t_1} + b), \quad x_2 = a \sin(\sqrt{t_2} + b). \]

Substitute \( \sqrt{t_1} + b = \frac{\pi}{2} + k\pi \) and \( \sqrt{t_2} + b = \frac{\pi}{2} + (k + 1)\pi \):

\[ x_1 = a \sin\left(\frac{\pi}{2} + k\pi\right), \quad x_2 = a \sin\left(\frac{\pi}{2} + (k + 1)\pi\right). \]

Using the property of sine:

\[ \sin\left(\frac{\pi}{2} + k\pi\right) = (-1)^k, \quad \sin\left(\frac{\pi}{2} + (k + 1)\pi\right) = (-1)^{k + 1}. \]

Thus:

\[ x_1 = a(-1)^k, \quad x_2 = a(-1)^{k + 1}. \]

The distance between the two points is:

\[ \text{Distance} = |x_2 - x_1| = |a(-1)^{k + 1} - a(-1)^k|. \]

Since \( (-1)^{k + 1} - (-1)^k = -2(-1)^k \), the absolute value gives:

\[ \text{Distance} = 2a. \]

Conclusion: The particle comes to rest at two points whose distance is:

\[ 2a. \]