Question

The sound level in decibels is typically expressed as \(\beta=10 \log \left(I / I_{0}\right),\) but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on the amplitude squared, so the expression is \(\beta=20 \log \left(P / P_{0}\right),\) where \(P_{0}\) is the smallest pressure difference noticeable by the ear: \(P_{0}=2.00 \cdot 10^{-5} \mathrm{~Pa}\). A loud rock concert has a sound level of \(110 . \mathrm{dB}\), find the amplitude of the pressure wave generated by this concert.

Short answer

Answer: The amplitude of the pressure wave generated by the rock concert is approximately \(1.58 \mathrm{Pa}\).

Step-by-step solution

Write down the given information

We have the following information: - Sound level at the rock concert, \(\beta = 110\mathrm{dB}\) - Smallest pressure difference noticeable by the ear, \(P_{0} = 2.00 \cdot 10^{-5} \mathrm{Pa}\)

Write down the expression for sound level in decibels

The expression for sound level in decibels (\(\beta\)) is given by: $$ \beta=20 \log \left(\frac{P} {P_{0}}\right) $$

Plug in the given information into the expression for \(\beta\)

We will plug in the given values of \(\beta\) and \(P_{0}\) into the expression for \(\beta\): $$ 110 = 20 \log \left(\frac{P} {2.00 \cdot 10^{-5}}\right) $$

Solve for \(P\)

Now, we need to solve this equation for \(P\). Follow these steps: 1. Divide both sides by 20: $$ 5.5 = \log \left(\frac{P} {2.00 \cdot 10^{-5}}\right) $$ 2. Apply anti-logarithm to eliminate the logarithm: $$ 10^{5.5} = \frac{P} {2.00 \cdot 10^{-5}} $$ 3. Multiply both sides by \(2.00 \cdot 10^{-5}\): $$ P = 10^{5.5} \times 2.00 \cdot 10^{-5} \mathrm{Pa} $$ 4. Calculate the value of \(P\): $$ P \approx 1.58 \mathrm{Pa} $$ Thus, the amplitude of the pressure wave generated by the rock concert is approximately \(1.58 \mathrm{Pa}\).

Key concepts

Decibels

Decibels are a logarithmic unit used to measure the intensity of sound, represented by the symbol dB. This unit helps us understand sounds in a way that's easy to interpret across a wide range of intensity levels, from a whisper to a loud rock concert. The reason for using a logarithmic scale is that the human ear perceives sound intensity logarithmically rather than linearly.
The basic formula for sound level in decibels, when knowing the intensity, is:

• \(\beta = 10 \log \left(\frac{I}{I_{0}}\right)\)

Here, \(I\) is the sound intensity, and \(I_{0}\) is the reference intensity, which is typically the threshold of hearing. However, since sound is conveyed through changes in pressure, we often use pressure difference in calculations. For pressure, the formula becomes:

• \(\beta = 20 \log \left(\frac{P}{P_{0}}\right)\)

In this case, \(P\) is the pressure amplitude, and \(P_{0}\) is the smallest pressure noticeable by our ears.

Pressure Wave

Sound travels through the air as a pressure wave, which means areas of high pressure alternate with areas of low pressure. This oscillation creates vibrations that are perceived as sound by our ears. The amplitude of a pressure wave relates to how "strong" the sound is and is crucial in determining the sound level (or loudness) in decibels.
At a rock concert, for instance, sound waves with higher amplitudes create louder sounds. The smallest pressure difference that the human ear typically notices is about \(2.00 \cdot 10^{-5} \mathrm{~Pa}\). This is a very tiny pressure, showing just how sensitive our hearing is. When dealing with these small amplitudes, comparing them through logarithmic decibels makes it easier to deal with and understand large variations in sound intensity.

Logarithmic Expressions

Logarithmic expressions simplify the comparison of very large or very tiny numbers by transforming them into manageable figures. When sound is defined in decibels, we use logarithmic expressions to encapsulate how the intensity or pressure of a sound wave compares to a reference level.
The math uses base 10 logarithms. Here's what happens when determining sound level:

• First, find the ratio of current pressure \(P\) to the reference pressure \(P_{0}\).
• Then apply the logarithm to this ratio \(\log \left(\frac{P}{P_{0}}\right)\).
• Multiply the result by 20, which adjusts the scale to decibels.

Using a log-based scale means that each increase of 10 dB represents a tenfold increase in sound intensity. So, a small change in decibels actually represents a large change in intensity or pressure of sound waves. This highlights the power of logarithmic expressions in handling real-world phenomena like sound.