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This is one of our many quizzes on Hearing. Click on this to access more such competitive quizzes on Sound.

Hearing Quiz 2

physics

 Two people are standing with respect to a sound source such that the intensities of the sound for them are $7{{I}_{o}}$ and $21{{I}_{o}}$ respectively. What is the difference in their respective sound levels? 

 The sound level is directly proportional to the intensity of the sound so it is also called intensity level. 

The sound level of a sound is given by; $\beta =10\log \left( \dfrac{I}{{{I}_{o}}} \right)$ Substituting the values; ${{\beta }_{1}}=10\log \left( \dfrac{{{I}_{1}}}{{{I}_{0}}} \right)=10\log \left( \dfrac{7{{I}_{0}}}{{{I}_{0}}} \right) \\ {{\beta }_{1}}=10\log \left( 7 \right)$ Similarly, ${{\beta }_{2}}=10\log \left( \dfrac{{{I}_{2}}}{{{I}_{0}}} \right)=10\log \left( \dfrac{21{{I}_{0}}}{{{I}_{0}}} \right) \\ {{\beta }_{2}}=10\log \left( 21 \right)$ The difference is given by; $\Delta \beta ={{\beta }_{2}}-{{\beta }_{1}} \\ \Rightarrow \Delta \beta =10\left[ \log \left( 21 \right)-\log \left( 7 \right) \right] \\ \Rightarrow \Delta \beta =10\left[ \log \left( \dfrac{21}{7} \right) \right]=10\log \left( 3 \right) \\ \therefore \Delta \beta =4.77\text{ dB}$

 A sound wave is travelling through a monatomic ideal gas. If the temperature of the gas increases, what happens to the sound level? 

 The volume of the medium will change with temperature and so will the density. Changing density affects velocity of the sound wave. 

The given medium is a monatomic ideal gas. So, it will obey Charles's Law of ideal gases. From the Charles’s Law; $V\propto T$ Here, $V$ is volume and $T$ is temperature. But density is given as; $\rho \propto \dfrac{1}{T}$ And velocity of the sound wave is given by; $v\propto \dfrac{1}{\sqrt{\rho }} \\ \Rightarrow v\propto \sqrt{T}$ Also, intensity increases with velocity as; $I\propto v \\ \Rightarrow I\propto \sqrt{T}$ And sound level depends on intensity as; $\beta \propto \log \left( I \right) \\ \therefore \beta \propto \log \left( \sqrt{T} \right)$

 Two sound sources $A$ and $B$ emit sounds with equal angular frequencies. The velocities of these sound waves are $2v$ and $8v$ respectively. Assuming that both the sources emit their sounds in the same medium, calculate the difference in their sound levels. 

 The intensity of the sound is directly proportional to the speed of the wave and the sound level is proportional to the log of intensity ratio.

Given that; ${{v}_{a}}=2v \\ \Rightarrow {{v}_{b}}=8v$ The intensities of the two sounds will be given by; ${{I}_{a}}=\dfrac{1}{2}\rho {{v}_{a}}{{\omega }^{2}} \\ \Rightarrow {{I}_{a}}=\dfrac{1}{2}2v\rho {{\omega }^{2}}=v\rho {{\omega }^{2}} \\ {{I}_{b}}=\dfrac{1}{2}\rho {{v}_{b}}{{\omega }^{2}} \\ \Rightarrow {{I}_{b}}=\dfrac{1}{2}8v\rho {{\omega }^{2}}=4\rho v{{\omega }^{2}} $ Sound levels will be given as; ${{\beta }_{a}}=10\log \left( \dfrac{{{I}_{a}}}{{{I}_{0}}} \right)=10\log \left( \dfrac{\rho v{{\omega }^{2}}}{{{10}^{-12}}} \right) \\ \Rightarrow {{\beta }_{a}}=10\left[ \log \left( \rho v{{\omega }^{2}} \right)-\log \left( {{10}^{-12}} \right) \right] \\ \Rightarrow {{\beta }_{a}}=10\left[ \log \left( \rho v{{\omega }^{2}} \right)+12 \right] \\ {{\beta }_{b}}=10\log \left( \dfrac{{{I}_{b}}}{{{I}_{0}}} \right)=10\log \left( \dfrac{4\rho v{{\omega }^{2}}}{{{10}^{-12}}} \right) \\ \Rightarrow {{\beta }_{b}}=10\left[ \log \left( 4\rho v{{\omega }^{2}} \right)-\log \left( {{10}^{-12}} \right) \right] \\ \Rightarrow {{\beta }_{b}}=10\left[ \log \left( 4\rho v{{\omega }^{2}} \right)+12 \right]$ The difference of sound levels is given by; $\Delta \beta ={{\beta }_{b}}-{{\beta }_{a}} \\ \Rightarrow \Delta \beta =10\left\{ \left[ \log \left( 4\rho v{{\omega }^{2}} \right)+12 \right]-\left[ \log \left( \rho v{{\omega }^{2}} \right)+12 \right] \right\} \\ \Rightarrow \Delta \beta =10\log \left( 4 \right) \\ \therefore \Delta \beta =6.02\text{ dB}$

 If a person hears a sound of $53\text{ dB}$ , calculate the speed of the sound wave if $\omega =\pi $ and density of the medium is $1.29\text{ kg}{{\text{m}}^{-3}}$. 

 The sound level of the sound wave is directly proportional to the log of the intensity ratio and the intensity is directly proportional to the speed of the wave. 

The sound level for a sound wave is given by; $\beta =10\log \left( \dfrac{I}{{{I}_{0}}} \right) \\ {{I}_{0}}={{10}^{-12}}\text{ W}{{\text{m}}^{-2}} \\ \Rightarrow 53=10\log \left( \dfrac{I}{{{10}^{-12}}} \right) \\ \Rightarrow 53=10\log I-10\log \left( {{10}^{-12}} \right) \\ \Rightarrow 53-120=10\log I \\ \Rightarrow \log I=\dfrac{-67}{10} \\ \Rightarrow I=2\times {{10}^{-7}}\text{ W}{{\text{m}}^{-2}}$ The intensity of the sound wave is given by; $I=\dfrac{1}{2}\rho v{{\omega }^{2}} \\ \Rightarrow v=\dfrac{2I}{\rho {{\omega }^{2}}} \\ \Rightarrow v=\dfrac{2\times 2\times {{10}^{-7}}}{1.29\times {{\pi }^{2}}} \\ \Rightarrow v=\dfrac{4\times {{10}^{-7}}}{12.73} \\ \therefore v=3.14\times {{10}^{-8}}\text{ m}{{\text{s}}^{-1}}$

 A hypothetical point sound source dissipates $400\text{ kJ}$ of energy for $8\text{ sec}$.Calculate the sound level of the sound heard by a man standing $40\text{ m}$ away from the source. 

 The intensity of the sound is the ratio of the power dissipated by the source by the surface area.The surface area will be sphere-like because the source is a point source. 

Power is given by; $P=Et \\ \Rightarrow P=400\times 8 \\ \Rightarrow P=3200\text{ kW} $ Area will be given as; $A=4\pi {{r}^{2}} \\ \Rightarrow A=4\pi \left( 40\times 40 \right) \\ \Rightarrow A=20106.19\text{ }{{\text{m}}^{2}} $ Intensity is given by; $I=\dfrac{P}{A} \\ \Rightarrow I=\dfrac{3200\times {{10}^{3}}\text{ W}}{20106.19\text{ }{{\text{m}}^{2}}} \\ \Rightarrow I=159.15\text{ W}{{\text{m}}^{-2}} $ The sound level is given by; $\beta =10\log \left( \dfrac{I}{{{I}_{0}}} \right) \\ {{I}_{0}}={{10}^{-12}} \\ \Rightarrow \beta =10\log \left( \dfrac{159.15}{{{10}^{-12}}} \right) \\ \Rightarrow \beta =10\left[ \log \left( 159.15 \right)-\log \left( {{10}^{-12}} \right) \right] \\ \Rightarrow \beta =10\left( 2.20+12 \right) \\ \therefore \beta =142.02\text{ dB}$