The intensity (in W/m2) of one sound is 4.83 times as great as the intensity of another sound. Relative to the quieter sound, what is the intensity level â of the louder sound?
I am not sure how to start this. Thank you.
db = 10 log intensity ratio
so I would do log of 4.83 then times 10?
To find the intensity level (in decibels, dB) of the louder sound relative to the quieter sound, you can use the formula:
L = 10 log(I/I₀)
Where:
L is the intensity level in decibels,
I is the intensity of the louder sound, and
I₀ is the intensity of the quieter sound.
In this case, you are given that the intensity of the louder sound is 4.83 times as great as the intensity of the quieter sound. Let's assume the intensity of the quieter sound as I₀. Then, the intensity of the louder sound would be 4.83 * I₀.
Plugging these values into the formula, we get:
L = 10 log((4.83 * I₀)/I₀)
Simplifying further:
L = 10 log(4.83)
Using a calculator, we find:
L ≈ 10 × 0.684
L ≈ 6.84 dB
Therefore, the intensity level (in decibels) of the louder sound relative to the quieter sound is approximately 6.84 dB.