Chapter 1, Problem 1P

A 2 kHz sound wave traveling in the x direction in air was observed to have a differential pressure p(x, t) = 10 N/m² at x = 0 and t = 50 μs. If the reference phase of p(x, t) is 36°, find a complete expression for p(x, t). The velocity of sound in air is 330 m/s.

To determine

The complete expression for differential pressure $p\left(x,t\right)$.

Answer to Problem 1P

The complete expression for $p\left(x,t\right)$ is,

$\underset{\_}{p\left(x,t\right)=32.36\cos \left(4\pi \times {10}^{3}t-38.08x+36°\right){\text{N/m}}^{\text{2}}}$.

Explanation of Solution

Given data:

The differential pressure $p\left(x,t\right)$ of a sound wave that is travelling in the $x$ direction in air is $10{\text{N/m}}^{\text{2}}$ at $x=0$ and $t=50\mu \text{s}$.

The reference phase of $p\left(x,t\right)$ is $36°$.

The velocity of the sound in air is $330\text{m/s}$.

The frequency of the sound wave is $2\text{kHz}$.

Calculation:

The formula to calculate the angular frequency is given as,

$\omega =2\pi f$

Here,

f is frequency of the sound wave and

$\omega$ is angular frequency.

Substitute $2\times {10}^3$ for $f$ to obtain angular frequency.

$\begin{array}{c}\omega =2\pi \left(2\times {10}^3\right)\\ =4\pi \times {10}^3\text{rad/s}\end{array}$

The formula for the wavelength of the sound wave is given as,

$\lambda =\frac{u_{\text{p}}}{f}$

Here,

$u_{\text{p}}$ is velocity of the sound in the air and

$\lambda$ is the wavelength.

Substitute $2\times {10}^3$ for $f$ and $330$ for $u_{\text{p}}$ in the equation to obtain the wavelength.

$\begin{array}{c}\lambda =\frac{330}{2\times {10}^3}\\ =0.165\text{m}\end{array}$

The formula for the phase constant is given as,

$\beta =\frac{2\pi }{\lambda }$

Here, the phase constant is $\beta$.

Substitute $0.165$ for $\lambda$ to obtain the phase constant.

$\begin{array}{c}\beta =\frac{2\pi }{0.165}\\ =38.08\end{array}$

The expression for the differential pressure is given as follows.

$p\left(x,t\right)=A\cos \left(\omega t-\beta x+{\varphi }_0\right)$ (1)

Here,

$p\left(x,t\right)$ is the differential pressure of a sound wave that is travelling in the $x$ direction in air,

${\varphi }_0$ is the reference phase of $p\left(x,t\right)$, and

$t$ is time.

Substitute $10{\text{N/m}}^{\text{2}}$ for $p\left(x,t\right)$, $4\pi \times {10}^3\text{rad/s}$ for $\omega$, $50\times {10}^{-6}$ for $t$, $38.08$ for $\beta$, $0$ for $x$ and $36°$ for ${\varphi }_0$ in equation (1).

$\begin{array}{c}10=A\cos \left(4\pi \times {10}^3\times 50\times {10}^{-6}-38.0799\times \left(0\right)+36°\right)\\ 10=A\cos \left(0.6283-0+36°\right)\\ 10=A\cos \left(0.6283\times \frac{180°}{\pi }+36°\right)\\ 10=A\cos \left(36°+36°\right)\end{array}$

Further solving the above expression as,

$\begin{array}{l}10=A\cos \left(72°\right)\\ A=32.36\end{array}$

Substitute $32.36$ for $A$, $4\pi \times {10}^3\text{rad/s}$ for $\omega$, $38.08$ for $\beta$, and $36°$ for ${\varphi }_0$ in equation (1).

$p\left(x,t\right)=32.36\cos \left(4\pi \times {10}^{3}t-38.08x+36°\right)$

Conclusion:

Therefore, the complete expression for $p\left(x,t\right)$ is,

$p\left(x,t\right)=32.36\cos \left(4\pi \times {10}^{3}t-38.08x+36°\right)$.