what about this solution: The sound wave is a plane wave traveling in the positive \(x\)-direction, so the general form of the differential pressure is: \[ p(x,t) = P_m \cos(\omega t - kx + \phi) \] where: - \(P_m\) is the amplitude, - \(\omega\) is the angular frequency, - \(k\) is the wave number, - \(\phi\) is the phase constant. Given: - Frequency \(f = 5 \text{kHz} = 5000 \text{Hz}\), - Speed of sound \(v = 330 \text{m/s}\), - At \(x = 0\), \(t = 45 \mu\text{s} = 45 \times 10^{-6} \text{s}\), \(p(0, 45 \times 10^{-6}) = 15 \text{N/m}^2\), - Reference phase of \(p(x,t)\) is \(25^\circ\), which is interpreted as the phase angle at \(x = 0\) and \(t = 45 \times 10^{-6} \text{s}\). ### Step 1: Calculate angular frequency \(\omega\) \[ \omega = 2\pi f = 2\pi \times 5000 = 10000\pi \text{rad/s} \] ### Step 2: Calculate wave number \(k\) \[ k = \frac{\omega}{v} = \frac{10000\pi}{330} = \frac{1000\pi}{33} \text{rad/m} \] ### Step 3: Determine the phase constant \(\phi\) The phase at any point is \(\theta = \omega t - kx + \phi\). At \(x = 0\), \(t = 45 \times 10^{-6} \text{s}\), \(\theta = 25^\circ\). First, compute \(\omega t\): \[ \omega t = 10000\pi \times (45 \times 10^{-6}) = 10000\pi \times 4.5 \times 10^{-5} = 0.45\pi \text{rad} \] Convert to degrees: \[ 0.45\pi \text{rad} \times \frac{180^\circ}{\pi} = 81^\circ \] Since \(x = 0\), the phase is: \[ \theta = \omega t + \phi = 81^\circ + \phi = 25^\circ \] Solving for \(\phi\): \[ \phi = 25^\circ - 81^\ci