The amplitude of a sinusoidal sound wave is quadrupled.
a) By which factor is its intensity multiplied?
b)By how many decibels does the sound intensity level increase?
Intensity is proportional to amplitude square (power is force times speed and both are doubled) so 4
10 log I2/I1 or 20 log A2/A1
20 log 2 = 6 Db
To answer these questions, we need to understand the relationships between amplitude, intensity, and sound intensity level.
a) The intensity of a sound wave is directly proportional to the square of its amplitude. Therefore, if the amplitude of a sinusoidal sound wave is quadrupled (increased by a factor of 4), the intensity will be multiplied by the square of that factor: 4^2 = 16. Hence, the intensity is multiplied by a factor of 16.
b) To determine how many decibels the sound intensity level increases, we can use the formula:
ΔL = 10 * log10(I2 / I1)
ΔL represents the change in sound intensity level, I2 is the final intensity, and I1 is the initial intensity.
Since the intensity increased by a factor of 16 (as determined in part a), the final intensity (I2) is 16 times the initial intensity (I1). Therefore, ΔL can be calculated as:
ΔL = 10 * log10(16 / 1) = 10 * log10(16) ≈ 10 * 1.204 = 12.04 decibels.
Hence, the sound intensity level increases by approximately 12.04 decibels.