Question

Position of a particle is given by \( x = A \sin \left( 50t + \frac{\pi}{3} \right) \). If speed and acceleration become 0 for the first time at \( t_1 \) and \( t_2 \) sec respectively, then find \( t_1 \) and \( t_2 \) (in sec):

• A. \( \frac{\pi}{300}, \frac{\pi}{75} \)
• B. \( \frac{\pi}{300}, \frac{\pi}{150}
• C. \( \frac{\pi}{150}, \frac{\pi}{75}
• D. \( \frac{\pi}{150}, \frac{\pi}{300}

Hint:

To find the time at which speed or acceleration becomes zero in simple harmonic motion, differentiate the position equation to obtain speed and acceleration, then solve for when these derivatives are zero.

Answer 1

The Correct Option is A

Step 1: Position, Velocity, and Acceleration.
The displacement of the particle is given by: \[ x = A \sin \left( 50t + \frac{\pi}{3} \right) \] The velocity \( v \) is obtained by differentiating the displacement with respect to time: \[ v = \frac{dx}{dt} = 50A \cos \left( 50t + \frac{\pi}{3} \right) \] The acceleration \( a \) is the derivative of velocity with respect to time: \[ a = \frac{dv}{dt} = -50^2 A \sin \left( 50t + \frac{\pi}{3} \right) \]
Step 2: Determining the time when velocity becomes zero (\( t_1 \)).
Velocity becomes zero when the cosine term is zero: \[ \cos \left( 50t + \frac{\pi}{3} \right) = 0 \] The general solution is: \[ 50t + \frac{\pi}{3} = \frac{\pi}{2} + n\pi \] Solving for \( t \): \[ 50t = \frac{\pi}{2} + n\pi - \frac{\pi}{3} \] \[ 50t = \frac{3\pi}{6} - \frac{2\pi}{6} + n\pi \] \[ 50t = \frac{\pi}{6} + n\pi \] \[ t_1 = \frac{\pi}{300} + \frac{n\pi}{50} \] The smallest positive value occurs when \( n = 0 \): \[ t_1 = \frac{\pi}{300} \, \text{sec} \]
Step 3: Determining the time when acceleration becomes zero (\( t_2 \)).
Acceleration becomes zero when the sine term is zero: \[ \sin \left( 50t + \frac{\pi}{3} \right) = 0 \] The general solution is: \[ 50t + \frac{\pi}{3} = n\pi \] Solving for \( t \): \[ 50t = n\pi - \frac{\pi}{3} \] \[ t_2 = \frac{n\pi}{50} - \frac{\pi}{150} \] The smallest positive value occurs when \( n = 1 \): \[ t_2 = \frac{\pi}{75} \, \text{sec} \]
Final Answer: \( t_1 = \frac{\pi}{300} \, \text{sec}, \quad t_2 = \frac{\pi}{75} \, \text{sec} \).