Question

Question: The pressure in a traveling sound wave is given by the equation

.$\mathit{\Delta }\mathbf{p}\mathbf{=}\left(1.50Pa\right)\mathit{s}\mathit{i}\mathit{n}\mathit{\pi }\left[\left(1.900m^{-1}\right)x-\left(1315s^{-1}\right)t\right]$

Find the (a) Pressure amplitude, (b) Frequency, (c) Wavelength, and (d) Speed of the wave

Short answer

Answer

1. Pressure amplitude is 1.50 Pa.
2. Sound wave has frequency 158 Hz.
3. Sound wave has wavelength2.22 m .
4. Velocity of the sound wave is 350 m/s.

Step-by-step solution

Given

The change in pressure of sound wave is,

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

Determining the concept

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.

(a) Determining the

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,

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

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

(b) Determining the

Comparing equation (1) and (2),

$\omega =315\pi \text{rad/s}$

and using,

$$\begin{gathered} f=\frac{\omega }{2\pi } \\ f=\frac{315\pi \text{rad/s}}{2\times \pi } \\ f=15\text{8Hz} \end{gathered}$$

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

(c) Determining the

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

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

Hence, sound wave has wavelength 2.22m .

(d) Determining the

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

$$\begin{gathered} v=\lambda f \\ =2.22\text{m}\times 158\text{Hz} \\ =3\text{50m/s} \end{gathered}$$

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