If sound is increased in intensity by a factor of 100,000, what is the change in decibels?
I thought 50?
other options are 5, 500, and 100,000. Please explain because I'm confused on how this works.
I thought 50 dB as well.
L(dB) = 10 log₁₀(P₁/P₀)= 10 log₁₀100000= =10•5=50
Thanks very much guys!!
To determine the change in decibels when the intensity of sound is increased by a factor of 100,000, we need to use the logarithmic formula for the decibel scale:
Change in Decibels (dB) = 10 * log10 (Intensity2 / Intensity1)
To find the change in decibels, we need to compare the new and original intensities. Let's assign some values to make it easier to understand:
Let Intensity1 be the original intensity of sound.
Let Intensity2 be the new intensity after it is increased by a factor of 100,000.
The formula becomes:
Change in Decibels (dB) = 10 * log10 (Intensity2 / Intensity1)
Now, let's calculate the change in decibels:
Change in Decibels (dB) = 10 * log10 (Intensity2 / Intensity1)
Change in Decibels (dB) = 10 * log10 (100,000 * Intensity1 / Intensity1)
Change in Decibels (dB) = 10 * log10 (100,000)
Change in Decibels (dB) = 10 * log10 (10^5)
Change in Decibels (dB) = 10 * 5
Change in Decibels (dB) = 50
Therefore, the change in decibels is 50.
In your initial thought, you guessed 50, which is correct. Options 5, 500, and 100,000 are incorrect because they do not account for the logarithmic nature of the decibel scale. The decibel scale compresses large changes in sound intensity into smaller numerical values by using logarithmic calculations.