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Question: The amplitude of motion for the air in the path of \(60dB\), \(800Hz\) a sound wave is \(1.36 \times...

The amplitude of motion for the air in the path of 60dB60dB, 800Hz800Hz a sound wave is 1.36×10x1.36 \times {10^x}. Find xx Assume that ρair=1.29kg/m3{\rho _{air}} = 1.29kg/{m^3} and v=330m/sv = 330m/s.

Explanation

Solution

We need to write the expression for loudness and using that we need to obtain the value of intensity. We know the relation between amplitude and intensity, so using that we can arrive at our desired result.

Complete Step-By-Step Solution:
In the question, it is given that, loudness, L=6dBL = 6dB
We know, the formula of loudness, can be written as:
L=log⁡(IIo)L = \log (\dfrac{I}{{{I_o}}})
Where:
I=I = Intensity
Io={I_o} = Reference Intensity
Loudness of sound relates the intensity of the heard sound with the intensity of sound at threshold of hearing. This is also known as reference intensity.
Intensity of sound on the other hand is power of sound measured in watts divided by the area that the sound covers, measured in square-meters.
We know, the value of reference intensity is Io=10−12W/m2{I_o} = {10^{ - 12}}W/{m^2}
Putting the values, in the equation of loudness, we can write:
IIo=106\dfrac{I}{{{I_o}}} = {10^6}
On solving this, we obtain:
I=10−6W/m2I = {10^{ - 6}}W/{m^2}
We know, intensity of sound can also be written as:
I=2π2f2A2ρvI = 2{\pi ^2}{f^2}{A^2}\rho v
Where:
f=f = Frequency
v=v = Velocity of sound
ρ=\rho = Density of air
On rearranging the equation, we get:
I2π2f2ρv=A2\dfrac{I}{{2{\pi ^2}{f^2}\rho v}} = {A^2}
Now, putting the values, as given in the question, we get:
10−62(3.14)2(800)2(1.29)(330)=A2\dfrac{{{{10}^{ - 6}}}}{{2{{(3.14)}^2}{{(800)}^2}(1.29)(330)}} = {A^2}
This comes out to be:
A=1.36×10−8mA = 1.36 \times {10^{ - 8}}m
In the question, it is given that the amplitude of the sound is =1.36×10xm = 1.36 \times {10^x}m
Therefore. On equating the two values, we obtain:
x=−8x = - 8
This is the required solution.

Note:
When we say the loudness of a sound increases, we mean that the intensity of sound in dBdB increases. With increases in intensity, it actually implies that the pressure of the wave increases, thus we obtain louder sounds at higher intensities. As loudness of the sound increases its amplitude also increases.