The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound level goes up by 38.0 dB?
20*Log(V1/Vo) = 38 db
Log(V1/Vo) = 1.9
V1/Vo = 79.4
To determine the factor by which the amplitude of a sound wave increases when the sound level goes up by a certain amount, we need to understand the relationship between sound level and amplitude.
Sound level is measured in decibels (dB), and it is a logarithmic scale that quantifies the intensity or loudness of a sound relative to a reference level. The relationship between sound level and amplitude is given by the following equation:
Sound level (L) = 10 * log10(A^2 / A_ref^2)
Where:
- L is the sound level in decibels (dB)
- A is the amplitude of the sound wave
- A_ref is the reference amplitude (typically 20 μPa for sound in air)
Now, let's find the factor by which the amplitude increases when the sound level goes up by 38.0 dB.
First, let's rearrange the equation to solve for A:
L = 10 * log10(A^2 / A_ref^2)
L/10 = log10(A^2 / A_ref^2)
10^(L/10) = A^2 / A_ref^2
Next, let's express the increase in sound level (38.0 dB) in terms of the initial and final sound levels:
ΔL = L_final - L_initial
38.0 = L_final - L_initial
Since the factor by which the amplitude increases is related to the change in sound level, we can use:
Factor = 10^(ΔL/10)
Plugging in the given value for ΔL into the equation, we have:
Factor = 10^(38.0/10)
Calculating this expression, we find that the factor is approximately 63.0957.
Therefore, the amplitude of the sound wave increases by a factor of approximately 63.0957 when the sound level goes up by 38.0 dB.