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The change in pressure of sound wave is,

$螖\mathrm{p}=(1.50\mathrm{Pa})sin蟺[(1.900{\mathrm{m}}^{-1})\mathrm{x}-(1315{\mathrm{s}}^{-1})\mathrm{t}]$

By comparing the given equation of pressure variation with the general form of pressure variation of a sound wave, calculate the required values.

The formula is as follows:

$鈭�P\left(x,t\right)=鈭�P_m\sin \left(kx-wt\right)路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路\left(1\right)$

where P is pressure, k is spring constant, x is displacement,tis time and w is the angular frequency.

The change in pressure of sound wave is given as,

$鈭�P\left(x,t\right)=鈭�P_m\sin \left(kx-wt\right)路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路\left(1\right)$

The given pressure variation equation is,

$$\begin{gathered} 鈭�P=\left(1.50\text{pa}\right)\sin 蟺\left[\left(0.900{\text{m}}^{-1}\right)x-\left(315{\text{s}}^{-1}\right)t\right] \\ =\left(1.50\text{pa}\right)\sin \left[\left(0.900{\text{m}}^{-1}\right)蟺x-\left(315{\text{s}}^{-1}\right)蟺t\right]鈥�路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路\left(2\right) \end{gathered}$$

Comparing equation 2 with equation 1, the pressure amplitude of a sinusoidal wave is,

$P_m=1.\text{50Pa}$

Hence, pressure amplitude is 1.50 Pa

Comparing equation (1) and (2),

$蝇=315蟺\text{rad/s}$

and using,

$$\begin{gathered} f=\frac{蝇}{2蟺} \\ f=\frac{315蟺\text{rad/s}}{2脳蟺} \\ f=15\text{8Hz} \end{gathered}$$

Hence, sound wave has frequency,f = 158 Hz .

Comparing equations 1 and 2, $k=0.9蟺{\text{m}}^{-1}$and using,

$$\begin{gathered} 位=\frac{2蟺}{k} \\ =\frac{2蟺}{0.9蟺{\text{m}}^{-1}} \\ =2.\text{22m} \end{gathered}$$

Hence, sound wave has wavelength 2.22m .

To calculate the speed of the wave using the values of wavelength and wave number in,

$$\begin{gathered} v=位f \\ =2.22\text{m}脳鈥�158\text{Hz} \\ =3\text{50m/s} \end{gathered}$$

Hence, velocity of sound wave is 350m/s.