Question

the speed of sound in water is approximately 1480 m/s. write the equation, y(x,t). for a sound wave with a frequency of 1024 hz and an amplitude of 2.5 db travelling through water. explicitly show all calculations needed to determine the parameters of the equation - perhaps wavelength, wave number, period or natural frequency. in other words, whatever values you use in the equation, we need to see how you obtained them.

Explanation:

Step1: Calculate the wavelength

The formula for the speed of a wave is \(v = f\lambda\), where \(v\) is the speed, \(f\) is the frequency, and \(\lambda\) is the wavelength. Rearranging for \(\lambda\), we get \(\lambda=\frac{v}{f}\).
Substituting \(v = 1480\space m/s\) and \(f=1024\space Hz\), we have \(\lambda=\frac{1480}{1024}\approx1.445\space m\).

Step2: Calculate the wave number

The wave number \(k\) is given by \(k=\frac{2\pi}{\lambda}\).
Substituting \(\lambda\approx1.445\space m\), we get \(k = \frac{2\pi}{1.445}\approx4.34\space m^{-1}\).

Step3: Calculate the angular frequency

The angular frequency \(\omega\) is given by \(\omega = 2\pi f\).
Substituting \(f = 1024\space Hz\), we get \(\omega=2\pi\times1024\approx6434\space rad/s\).

Step4: Write the wave equation

The general form of a wave equation is \(Y(x,t)=A\sin(kx-\omega t)\).
Substituting \(A = 2.5\space dB\), \(k\approx4.34\space m^{-1}\), and \(\omega\approx6434\space rad/s\), we get \(Y(x,t)=2.5\sin(4.34x - 6434t)\).

Answer:

\(Y(x,t)=2.5\sin(4.34x - 6434t)\)