The equation of a sound wave in air is given by $p = (0.02\ \mathrm{Nm^{-2}})[\sin\,(500\ \mathrm{s^{-1}}\…
The equation of a sound wave in air is given by
$p = (0.02\ \mathrm{Nm^{-2}})[\sin\,(500\ \mathrm{s^{-1}}\ t) - (3\ \mathrm{m^{-1}})x]$
(i) Find the frequency, wavelength and the speed of sound wave in air.
(ii) If the equilibrium pressure of air is $1.01\times 10^5\ \mathrm{Nm^{-2}}$, what are the maximum and minimum pressure at a point as the wave passes through that point?
Solution
Sol. The given equation is
$p = 0.02\sin(500t - 3x)$ ...(i)
The standard equation is $p = p_0\sin(\omega t - kx)$ ...(ii)
Comparing Eqs. (i) and (ii), we get
$p_0 = 0.02\ \mathrm{Nm^{-2}},\ \omega = 500\ \mathrm{rad\ s^{-1}},\ k = 3\ \mathrm{m^{-1}}$
(i) Therefore, $f = \frac{\omega}{2\pi} = \frac{500}{2\pi} = \frac{250}{\pi}$ Hz
$k = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{k} = \frac{2\pi}{3}$ m
and $v = f\lambda = \frac{250}{\pi} \times \frac{2\pi}{3} = \frac{500}{3}\ \mathrm{ms^{-1}}$
(ii) Here, pressure, $p_{\max} = p'_0 + p_0$
$\quad\quad\quad\;= (1.01\times 10^5 + 0.02)\ \mathrm{Nm^{-2}}$
$\quad\quad\quad\;= 101000.02\ \mathrm{Nm^{-2}}$
and $\;p_{\min} = p'_0 - p_0 = (1.01\times 10^5 - 0.02)\ \mathrm{Nm^{-2}}$
$\quad\quad\quad\;= 100999.98\ \mathrm{Nm^{-2}}$
Here, $p'_0 = $ atmospheric pressure
Answer: $100999.98\ \mathrm{Nm^{-2}}$