91Ó°ÊÓ

The change in pressure of sound wave is,

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

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:

$∆P\left(x,t\right)=∆P_m\sin \left(kx-wt\right)······································\left(1\right)$

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,

$∆P\left(x,t\right)=∆P_m\sin \left(kx-wt\right)······································\left(1\right)$

The given pressure variation equation is,

$$\begin{gathered} ∆P=\left(1.50\text{pa}\right)\sin \mathrm{π}\left[\left(0.900{\text{m}}^{-1}\right)x-\left(315{\text{s}}^{-1}\right)t\right] \\ =\left(1.50\text{pa}\right)\sin \left[\left(0.900{\text{m}}^{-1}\right)\mathrm{π}x-\left(315{\text{s}}^{-1}\right)\mathrm{π}t\right] ····································\left(2\right) \end{gathered}$$

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),

$\mathrm{Ó\neg }=315\mathrm{Ï€}\text{rad/s}$

and using,

$$\begin{gathered} f=\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}} \\ f=\frac{315\mathrm{Ï€}\text{rad/s}}{2×\mathrm{Ï€}} \\ f=15\text{8Hz} \end{gathered}$$

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

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

$$\begin{gathered} \mathrm{λ}=\frac{2\mathrm{π}}{k} \\ =\frac{2\mathrm{π}}{0.9\mathrm{π}{\text{m}}^{-1}} \\ =2.\text{22m} \end{gathered}$$

Hence, sound wave has wavelength 2.22m .

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

$$\begin{gathered} v=\mathrm{λ}f \\ =2.22\text{m}× 158\text{Hz} \\ =3\text{50m/s} \end{gathered}$$

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