The provided wave equation is \( y = 2 \sin \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \). To determine acceleration, \( y \) must be differentiated twice with respect to time \( t \). The velocity \( v \) is the first derivative: \( v = \frac{\partial y}{\partial t} = 12\pi \cos \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \). The acceleration \( a \) is the second derivative: \( a = \frac{\partial^2 y}{\partial t^2} = -36\pi^2 \sin \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \). Substituting \( t = 1 \) and \( x = 4 \) into the acceleration equation yields: \( a = -36\pi^2 \sin \left( 2\pi \left( 3 \cdot 1 - 4 \right) + \frac{\pi}{4} \right) = -36\pi^2 \sin \left( -2\pi + \frac{\pi}{4} \right) = -36\pi^2 \sin \left( -\frac{7\pi}{4} \right) \). Evaluating this expression gives \( a = -36\pi^2 \cdot (-\frac{\sqrt{2}}{2}) = -36 \, \text{cm}^2/\text{s}^2 \). Therefore, the acceleration is \( -36 \, \text{cm}^2/\text{s}^2 \).