NTSE physics question Decibel (dB) is a unit of .loudness of sound. It is defined in a manner such that when amplitude of sound is multiplied by a factor of √10 , the: decibel level increases by 10 units. Loud music of 70 dB is being played at a function. To reduce the loudness to a level of 30 dB, the amplitude of the instrument playing music has to be reduced by a factor of 1. 10 2. 10√10 3. 100 Please explain! Thanks
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sound loudness in decibel = dB = 120 + 10 * Log_10 I_s
I_s = intensity of sound. logarithm to base 10.
If amplitude is multiplied by a factor of √10, then the intensity is multiplied by (√10)² = 10.
So loudness in dB' = 120 + 10 * [Log (√10)² + Log Is] = 10 + dB
So if 70 dB loud music is played at a function, if it to be reduced to 30 dB, its amplitude is reduced by a factor of n.
then dB' = 120 + 10 * Log ( I_s/ n²) = 120 + 10 * Log I_s - 10 Log n²
= dB - 20 Log n
dB - dB' = 20 Log n
70 - 30 = 20 log n
n = 10² = 100
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as decibel level changed by 40 dB, ie., 4 times 10 dB, so intensity changes by 10^4. Amplitude changes as square root of intensity, so 10².
I_s = intensity of sound. logarithm to base 10.
If amplitude is multiplied by a factor of √10, then the intensity is multiplied by (√10)² = 10.
So loudness in dB' = 120 + 10 * [Log (√10)² + Log Is] = 10 + dB
So if 70 dB loud music is played at a function, if it to be reduced to 30 dB, its amplitude is reduced by a factor of n.
then dB' = 120 + 10 * Log ( I_s/ n²) = 120 + 10 * Log I_s - 10 Log n²
= dB - 20 Log n
dB - dB' = 20 Log n
70 - 30 = 20 log n
n = 10² = 100
==================
as decibel level changed by 40 dB, ie., 4 times 10 dB, so intensity changes by 10^4. Amplitude changes as square root of intensity, so 10².
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