maximum - pressure amplitude that the human ear can tolerate in loud sounds is about $delta p = 28 pa$…

maximum - pressure amplitude that the human ear can tolerate in loud sounds is about $delta p = 28 pa$ (which is very much less than the normal air - pressure of about $p_0 = 10^5 pa$). a) what is the displacement amplitude for such a sound in air of density $\rho=1.2 kg/m^3$ at a frequency of $f = 1000.0 hz$ and speed $v = 343 m/s$? b) write down the wave function of sound wave for pressure amplitude? c) write down the wave function of sound wave for displacement amplitude? a) $delta p(x,t)=delta p_0sin(kx - omega t)$ $delta s(x,t)=delta s_0cos(kx - omega t)$ $omega = 2pi f$ $delta s_0=\frac{delta p_0}{\rho vomega}=\frac{28 n/m^2}{1.2 kg/m^3\times343 m/s\times2pi\times1000 hz}$ $m$ b) $delta p(x,t)=28 pasin(\frac{2pi f}{v}cdot x - 2pi fcdot t)$ c) $delta s(x,t)=delta s_0cos(\frac{2pi f}{v}cdot x - 2pi fcdot t)$ $p_0 = 1.01\times10^5 pa$ $\frac{kg}{m^3}cdot\frac{m}{s}cdot\frac{1}{s}$ $v = flambdarightarrowlambda=\frac{v}{f}$ $k = \frac{2pi}{lambda}=\frac{2pi f}{v}$ $k=\frac{2pi\times1000}{343}approx v$ $omega = 2pi f = 2pi\times1000$ $p_m = s_m\rho vomega$

maximum - pressure amplitude that the human ear can tolerate in loud sounds is about $delta p = 28 pa$ (which is very much less than the normal air - pressure of about $p_0 = 10^5 pa$). a) what is the displacement amplitude for such a sound in air of density $\rho=1.2 kg/m^3$ at a frequency of $f = 1000.0 hz$ and speed $v = 343 m/s$? b) write down the wave function of sound wave for pressure amplitude? c) write down the wave function of sound wave for displacement amplitude? a) $delta p(x,t)=delta p_0sin(kx - omega t)$ $delta s(x,t)=delta s_0cos(kx - omega t)$ $omega = 2pi f$ $delta s_0=\frac{delta p_0}{\rho vomega}=\frac{28 n/m^2}{1.2 kg/m^3\times343 m/s\times2pi\times1000 hz}$ $m$ b) $delta p(x,t)=28 pasin(\frac{2pi f}{v}cdot x - 2pi fcdot t)$ c) $delta s(x,t)=delta s_0cos(\frac{2pi f}{v}cdot x - 2pi fcdot t)$ $p_0 = 1.01\times10^5 pa$ $\frac{kg}{m^3}cdot\frac{m}{s}cdot\frac{1}{s}$ $v = flambdarightarrowlambda=\frac{v}{f}$ $k = \frac{2pi}{lambda}=\frac{2pi f}{v}$ $k=\frac{2pi\times1000}{343}approx v$ $omega = 2pi f = 2pi\times1000$ $p_m = s_m\rho vomega$

Answer

Explanation:

Step1: Recall the formula for displacement amplitude

The formula for displacement amplitude $\Delta S_0$ in terms of pressure amplitude $\Delta P_0$, density $\rho$, wave - speed $v$ and angular frequency $\omega$ is $\Delta S_0=\frac{\Delta P_0}{\rho v\omega}$. Given $\omega = 2\pi f$, where $f = 1000\ Hz$, $\rho=1.2\ kg/m^3$, $v = 343\ m/s$ and $\Delta P_0=28\ Pa$.

Step2: Calculate the angular frequency

$\omega=2\pi f=2\pi\times1000\ rad/s$.

Step3: Substitute values into the displacement - amplitude formula

$\Delta S_0=\frac{\Delta P_0}{\rho v\omega}=\frac{28}{1.2\times343\times2\pi\times1000}$. [ \begin{align*} \Delta S_0&=\frac{28}{1.2\times343\times2\pi\times1000}\ &=\frac{28}{1.2\times343\times2000\pi}\ &\approx1.1\times 10^{-5}\ m \end{align*} ]

Step4: Write the pressure - wave function

The general form of the pressure - wave function is $\Delta P(x,t)=\Delta P_0\sin(kx-\omega t)$. Here, $k = \frac{\omega}{v}=\frac{2\pi f}{v}$ and $\omega = 2\pi f$. So $\Delta P(x,t)=28\sin\left(\frac{2\pi\times1000}{343}x - 2\pi\times1000t\right)$.

Step5: Write the displacement - wave function

The general form of the displacement - wave function is $\Delta S(x,t)=\Delta S_0\cos(kx - \omega t)$. Substituting $\Delta S_0\approx1.1\times 10^{-5}\ m$, $k=\frac{2\pi f}{v}$ and $\omega = 2\pi f$, we get $\Delta S(x,t)=1.1\times 10^{-5}\cos\left(\frac{2\pi\times1000}{343}x-2\pi\times1000t\right)$.

Answer:

a) The displacement amplitude $\Delta S_0\approx1.1\times 10^{-5}\ m$. b) The pressure - wave function is $\Delta P(x,t)=28\sin\left(\frac{2000\pi}{343}x - 2000\pi t\right)$. c) The displacement - wave function is $\Delta S(x,t)=1.1\times 10^{-5}\cos\left(\frac{2000\pi}{343}x-2000\pi t\right)$.