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Hint: Generally one might think that the sound intensity and loudness are the same but they are very different from each other. The sound intensity (I) is defined as power (p) upon Area (A) while the loudness is the ratio of intensity of a given sound upon the intensity at the threshold of hearing. Human ear can only hear sounds in a particular range of intensities. The threshold of human hearing has an intensity of about 120dB. Now, Define the formula for sound level and then establish a proper relation between the current sound level and the final sound level while applying logarithmic properties, Use: (log (a.b) = loga+logb).
Formula Used:
Here we have used the formula
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Where
${B_o}$= Sound level
I = Intensity of sound
${I_o}$= ${I_o} = {10^{ - 12}}W/{m^2}$
Given:
I=4I
Find:
Sound level (B) = ??
Complete step by step answer:
Find out the intensity and solve for the unknown
Here we know the formula of sound wave which is
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Here ${I_o}$ is given by
${I_o} = {10^{ - 12}}W/{m^2}$
Now we have been given
I=4I
So now, it becomes
$B = 10\log (\dfrac{{4I}}{{{I_o}}})$
Apply trigonometric property (log (a.b) =loga+logb)
B =10(log$\dfrac{I}{{{I_o}}}$) +10log4
Since${B_o} = 10\log (\dfrac{I}{{{I_o}}})$, So
B = ${B_o}$+$10 \times 0.60$
B = (${B_o}$+6) dB
$\therefore$The new sound level becomes option (3) approximately B = (${B_o}$+6) dB.
Additional Information:
The sound energy is defined by the energy (sound) flowing per second through an area (unit) which is held at 90° to the direction of sound wave is known as the intensity of sound. Sound level on the other hand is the measurement of sound vibration or noise using logarithmic functions and their properties
Note: Go step by step first define the term sound intensity and then differentiate between sound intensity and loudness. Apply the formula that relates sound level to sound intensity and solve by using the logarithmic properties.
Formula Used:
Here we have used the formula
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Where
${B_o}$= Sound level
I = Intensity of sound
${I_o}$= ${I_o} = {10^{ - 12}}W/{m^2}$
Given:
I=4I
Find:
Sound level (B) = ??
Complete step by step answer:
Find out the intensity and solve for the unknown
Here we know the formula of sound wave which is
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Here ${I_o}$ is given by
${I_o} = {10^{ - 12}}W/{m^2}$
Now we have been given
I=4I
So now, it becomes
$B = 10\log (\dfrac{{4I}}{{{I_o}}})$
Apply trigonometric property (log (a.b) =loga+logb)
B =10(log$\dfrac{I}{{{I_o}}}$) +10log4
Since${B_o} = 10\log (\dfrac{I}{{{I_o}}})$, So
B = ${B_o}$+$10 \times 0.60$
B = (${B_o}$+6) dB
$\therefore$The new sound level becomes option (3) approximately B = (${B_o}$+6) dB.
Additional Information:
The sound energy is defined by the energy (sound) flowing per second through an area (unit) which is held at 90° to the direction of sound wave is known as the intensity of sound. Sound level on the other hand is the measurement of sound vibration or noise using logarithmic functions and their properties
Note: Go step by step first define the term sound intensity and then differentiate between sound intensity and loudness. Apply the formula that relates sound level to sound intensity and solve by using the logarithmic properties.
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