At the same distance the noise of a sandblaster is 13 dB louder than a power drill. This means the the displacement amplitude of the sound waves from the sandblaster is how many times greater than that of the power drill if both sounds have the same frequency?
Oh, well isn't this a sound showdown? So, we have the noise of a sandblaster being 13 dB louder than a power drill. Now, let's dive into the deep end of humor and try to figure this out.
You see, decibels are a logarithmic scale, which means that every 10 dB increase corresponds to a sound wave amplitude increase of 10 times. So, if the sandblaster is 13 dB louder than the power drill, we can take a leap of math and assume that the displacement amplitude of the sound waves from the sandblaster is roughly 10 to the power of (13/10) times greater.
In numerical terms, that would be approximately...*drumroll*...about 19.9526 times greater!
So, if you were imagining sound waves as trampolines, the displacement amplitude of the sandblaster's trampoline would be almost 20 times higher than the power drill's trampoline. Quite the acrobatic feat, wouldn't you say?
To determine the displacement amplitude ratio between the sandblaster and the power drill, we need to convert the difference in decibels to a difference in sound pressure levels (SPL).
The formula to convert decibels to sound pressure level (SPL) is:
SPL = 20log10(p/p0)
Where:
- SPL is the sound pressure level in decibels,
- p is the sound pressure,
- p0 is the reference sound pressure (usually taken as 20 microPascals).
Let's assume that the SPL of the power drill is x dB. According to the given information, the SPL of the sandblaster is 13 dB louder than the power drill. Therefore, the SPL of the sandblaster is (x+13) dB.
Using the formula above, we can find the sound pressure levels:
(x+13) = 20log10(p_sandblaster/p0)
x = 20log10(p_powerdrill/p0)
Now, let's solve for the ratio of sound pressure levels:
(x+13) - x = 20log10(p_sandblaster/p0) - 20log10(p_powerdrill/p0)
13 = 20log10(p_sandblaster/p_powerdrill)
Divide both sides by 20:
13/20 = log10(p_sandblaster/p_powerdrill)
Next, convert the logarithmic equation into an exponential equation:
10^(13/20) = p_sandblaster/p_powerdrill
Simplify the left side:
p_sandblaster/p_powerdrill = √(10^(13/10))
Thus, the displacement amplitude of the sound waves from the sandblaster is √(10^(13/10)) times greater than that of the power drill if both sounds have the same frequency.
To find the ratio of the displacement amplitudes of the sound waves from the sandblaster and the power drill, we can use the formula:
\[ \text{Sound Intensity} = \text{Amplitude}^2 \]
Given that the noise of the sandblaster is 13 dB louder than the power drill, we know that:
\[ \text{Intensity}_{\text{sandblaster}} = \text{Intensity}_{\text{power drill}} \times 10^{\left(\frac{13}{10}\right)} \]
As sound amplitude is directly proportional to the square root of sound intensity, we can take the square root of both intensities:
\[ \text{Amplitude}_{\text{sandblaster}} = \sqrt{\text{Intensity}_{\text{sandblaster}}} \]
\[ \text{Amplitude}_{\text{power drill}} = \sqrt{\text{Intensity}_{\text{power drill}}} \]
Therefore, the ratio of the displacement amplitudes is:
\[ \frac{{\text{Amplitude}_{\text{sandblaster}}}}{{\text{Amplitude}_{\text{power drill}}}} = \frac{{\sqrt{\text{Intensity}_{\text{sandblaster}}}}}{{\sqrt{\text{Intensity}_{\text{power drill}}}}} \]
To compute this value, we need the specific intensity values or amplitudes of both the sandblaster and power drill sounds.
20*Log(Ns/Np) = 13.
Log(Ns/Np) = 13/20 = 0.65,
Ns/Np = 10^0.65 = 4.47.
Ns = 4.47Np.