91Ó°ÊÓ

At time \( t=0 \), the wave equation simplifies to \( y(x, 0) = A \cos(kx) \). This represents the displacement of the wave as a function of position \( x \).

The velocity \( v_y \) is the time derivative of the displacement \( y(x, t) \), i.e. \( v_y(x, t) = \frac{\partial}{\partial t} y(x, t) = A \omega \sin(kx - \omega t) \). At \( t=0 \), this becomes \( v_y(x, 0) = A \omega \sin(kx) \).

Acceleration \( a_y \) is the time derivative of velocity, thus \( a_y(x, t) = \frac{\partial}{\partial t} v_y(x, t) = -A \omega^2 \cos(kx - \omega t) \). At \( t=0 \), the expression is \( a_y(x, 0) = -A \omega^2 \cos(kx) \).

For each point given, calculate \( y(x, 0) \), \( v_y(x, 0) \), and \( a_y(x, 0) \) to determine the state of the particle. Brief descriptions are then based on these calculations:1. Point \( x=0 \): Displacement = \( A \), Velocity = 0, Acceleration = \(-A \omega^2 \). The particle is momentarily stationary and accelerating downwards.2. Point \( x=\pi/4k \): Displacement = \( A/\sqrt{2} \), Velocity = \( A \omega /\sqrt{2} \), Acceleration = \(-A \omega^2/\sqrt{2} \). The particle is moving upwards and slowing down.3. Point \( x=\pi/2k \): Displacement = 0, Velocity = \( A \omega \), Acceleration = 0. The particle has the maximum upward velocity and no acceleration.4. Point \( x=3\pi/4k \): Displacement = \(-A/\sqrt{2} \), Velocity = \( A \omega /\sqrt{2} \), Acceleration = \(A \omega^2/\sqrt{2} \). The particle is moving upwards and speeding up.5. Point \( x=\pi/k \): Displacement = \(-A \), Velocity = 0, Acceleration = \(A \omega^2 \). The particle is stationary and accelerating upwards.6. and 7. Similarly analyze points \( x=5\pi/4k, 3\pi/2k \) for their displacement, velocity, and acceleration, checking changes in direction and speed.8. Point \( x=7\pi/4k \): Displacement = \(-A/\sqrt{2} \), Velocity = \(-A \omega /\sqrt{2} \), Acceleration = \(A \omega^2/\sqrt{2} \). The particle is moving downwards and speeding down.