sound intensity and decibel scale\n$i_0 = 10^{-12} \\ w/m^2$ ⇒ lowest value humans are sensitive\n decibel…

sound intensity and decibel scale\n$i_0 = 10^{-12} \\ w/m^2$ ⇒ lowest value humans are sensitive\n decibel scale: $\\beta(db)$\n$\\beta(db)=10\\log(\\frac{i}{i_0})$ lowest\n$i = i_0 ⇒$\n$\\beta(db)=10\\log(\\frac{i_0}{i_0})$\n$\\beta(db)=0$\n$i (pain) ⇒ 1.0 \\ w/m^2$\n talk ⇒ 60 db\n siren ⇒ 100 db\n a) calculate the sound intensity level in decibels for a sound wave traveling in air at $t = 32^{circ}c$ and having a pressure amplitude of $\\delta p = 0.85$ pa. air density $\\rho=1.29\\ kg/m^3$.\n b) if an amplifier decreases the sound level of the sound waves by $\\delta\\beta = 20.0$ db, what is the ratio of the final intensity of the waves to their initial intensity?\n$i=\\frac{p_m^2}{2\\rho v}$, $v(0^{circ}c)=331\\ m/s$\n$v(30^{circ}c)=331\\sqrt{\\frac{273 + 32}{273}}=?$\n$⇒ i=\\frac{28^2(n/m^2)^2}{2\\times1.29\\ kg/m^3\\times?\\ m/s}$

sound intensity and decibel scale\n$i_0 = 10^{-12} \\ w/m^2$ ⇒ lowest value humans are sensitive\n decibel scale: $\\beta(db)$\n$\\beta(db)=10\\log(\\frac{i}{i_0})$ lowest\n$i = i_0 ⇒$\n$\\beta(db)=10\\log(\\frac{i_0}{i_0})$\n$\\beta(db)=0$\n$i (pain) ⇒ 1.0 \\ w/m^2$\n talk ⇒ 60 db\n siren ⇒ 100 db\n a) calculate the sound intensity level in decibels for a sound wave traveling in air at $t = 32^{circ}c$ and having a pressure amplitude of $\\delta p = 0.85$ pa. air density $\\rho=1.29\\ kg/m^3$.\n b) if an amplifier decreases the sound level of the sound waves by $\\delta\\beta = 20.0$ db, what is the ratio of the final intensity of the waves to their initial intensity?\n$i=\\frac{p_m^2}{2\\rho v}$, $v(0^{circ}c)=331\\ m/s$\n$v(30^{circ}c)=331\\sqrt{\\frac{273 + 32}{273}}=?$\n$⇒ i=\\frac{28^2(n/m^2)^2}{2\\times1.29\\ kg/m^3\\times?\\ m/s}$

Answer

Explanation:

Step1: Recall the speed - temperature formula

The speed of sound in air as a function of temperature $T$ (in Celsius) is given by $v = v_0\sqrt{\frac{273 + T}{273}}$, where $v_0=331\ m/s$. For $T = 32^{\circ}C$, we have $v=331\sqrt{\frac{273 + 32}{273}}=331\sqrt{\frac{305}{273}}\approx348.7\ m/s$.

Step2: Recall the formula for sound intensity in terms of pressure amplitude

The sound intensity $I$ is given by $I=\frac{\Delta p_{max}^2}{2\rho v}$, where $\Delta p_{max}$ is the pressure amplitude, $\rho$ is the air - density, and $v$ is the speed of sound. Given $\Delta p_{max}=0.85\ Pa$, $\rho = 1.29\ kg/m^3$, and $v\approx348.7\ m/s$, we have $I=\frac{(0.85)^2}{2\times1.29\times348.7}=\frac{0.7225}{2\times1.29\times348.7}\approx9.9\times10^{-4}\ W/m^2$.

Step3: Calculate the sound intensity level

The sound intensity level $\beta$ (in decibels) is given by $\beta = 10\log\left(\frac{I}{I_0}\right)$, where $I_0 = 10^{-12}\ W/m^2$. Substituting $I\approx9.9\times10^{-4}\ W/m^2$ into the formula, we get $\beta=10\log\left(\frac{9.9\times10^{-4}}{10^{-12}}\right)=10\log(9.9\times10^{8})\approx10\times8.996\approx90\ dB$.

Step4: Solve for the intensity ratio in part (b)

We know that $\Delta\beta=\beta_2-\beta_1 = 20\ dB$. Since $\beta = 10\log\left(\frac{I}{I_0}\right)$, we have $\beta_2-\beta_1=10\log\left(\frac{I_2}{I_0}\right)-10\log\left(\frac{I_1}{I_0}\right)=10\log\left(\frac{I_2}{I_1}\right)$. Given $\Delta\beta = 20\ dB$, then $20 = 10\log\left(\frac{I_2}{I_1}\right)$. Dividing both sides by 10, we get $\log\left(\frac{I_2}{I_1}\right)=2$. By the definition of the logarithm, $\frac{I_2}{I_1}=10^2 = 100$.

Answer:

a. The sound intensity level is approximately $90\ dB$. b. The ratio of the final intensity to the initial intensity of the waves is $100$.