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):
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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.
Step 1: Position, Speed, and Acceleration.
The position of the particle is given as:
\[
x = A \sin \left( 50t + \frac{\pi}{3} \right)
\]
The speed \( v \) is the first derivative of position with respect to time \( t \):
\[
v = \frac{dx}{dt} = A \cdot 50 \cos \left( 50t + \frac{\pi}{3} \right)
\]
The acceleration \( a \) is the derivative of speed with respect to time \( t \):
\[
a = \frac{dv}{dt} = -A \cdot 50^2 \sin \left( 50t + \frac{\pi}{3} \right)
\]
Step 2: Speed becomes zero at time \( t_1 \).
For speed to be zero, the cosine term must be zero.
\[
\cos \left( 50t + \frac{\pi}{3} \right) = 0
\]
The general solution for this equation is:
\[
50t + \frac{\pi}{3} = \frac{\pi}{2} + n\pi
\]
Solving for \( t \), we get:
\[
50t = \frac{\pi}{2} + n\pi - \frac{\pi}{3} = \frac{3\pi}{6} + n\pi - \frac{2\pi}{6} = \frac{\pi}{6} + n\pi
\]
Thus:
\[
t_1 = \frac{\pi}{300} + \frac{n\pi}{50}
\]
The smallest value of \( t_1 \) (for \( n = 0 \)) is:
\[
t_1 = \frac{\pi}{300} \, \text{sec}
\]
Step 3: Acceleration becomes zero at time \( t_2 \).
For acceleration to be zero, the sine term must be zero.
\[
\sin \left( 50t + \frac{\pi}{3} \right) = 0
\]
The general solution for this equation is:
\[
50t + \frac{\pi}{3} = n\pi
\]
Solving for \( t \), we get:
\[
50t = n\pi - \frac{\pi}{3}
\]
Thus:
\[
t_2 = \frac{n\pi}{50} - \frac{\pi}{150}
\]
The smallest value of \( t_2 \) (for \( n = 1 \)) is:
\[
t_2 = \frac{\pi}{75} \, \text{sec}
\]
Final Answer: \( t_1 = \frac{\pi}{300} \, \text{sec}, \, t_2 = \frac{\pi}{75} \, \text{sec}.
