Question

A certain sound level is increased by an additional \(30 \mathrm{~dB}\). Show that ( \(a\) ) its intensity increases by a factor of 1000 and (b) its pressure amplitude increases by a factor of 32 .

Short answer

By using the logarithmic property of decibels, when the sound level is increased by 30 dB: (a) the intensity does increase by a factor of 1000, and (b) the pressure amplitude also increases by an approximately a factor of 32.

Step-by-step solution

Establishing the Decibel Formula

Begin by applying the formula of decibel scale, which is: \(L_{dB} = 10 \log_{10} \left(\frac{I}{I_0}\right)\)

Deriving the Intensity Factor

Next, in order to prove the claim (a), the decibel increase is given as 30 dB. By substituting in the formula, the equation can be written as \(30 = 10 \log_{10} \left(\frac{I}{I_0}\right)\). Rearranging the equation for \(I\) and \(I_0\), you get \(\frac{I}{I_0} = 10^{\frac{30}{10}}\), which simplifies to \(\frac{I}{I_0} = 10^3\) or \(\frac{I}{I_0} = 1000\). Therefore, the intensity increases by a factor of 1000.

Deriving the Pressure Amplitude Factor

To prove claim (b), knowing that the intensity is directly proportional to the square of the pressure amplitude \(I\propto P_{amplitude}^2\), the pressure amplitude increase may be written as \(\sqrt{\frac{I}{I_0}} = \frac{P_{amplitude}}{P_{initial}}\). Substituting \(\frac{I}{I_0} = 1000\) into the equation you get \(\sqrt{1000} =\frac{P_{amplitude}}{P_{initial}}\), which simplifies to \(\frac{P_{amplitude}}{P_{initial}} = 31.62\). Therefore, the pressure amplitude increases by an approximately a factor of 32.

Key concepts

Decibel Scale

The decibel scale is a logarithmic way of expressing sound intensity. This scale helps us understand how the intensity of a sound relates to what we can perceive. Since the human ear can hear sounds over a vast range of intensities, from the faintest rustle to a jet engine, using a linear scale isn't feasible. Instead, we use a logarithmic scale called the decibel scale, where each 10 dB increase typically represents a tenfold increase in intensity.

The formula for calculating decibels is:

• \( L_{dB} = 10 \log_{10} \left( \frac{I}{I_0} \right) \)

where:

• \( L_{dB} \) is the sound level in decibels,
• \( I \) is the intensity of the sound,
• \( I_0 \) is the reference intensity, typically the threshold of hearing (\( 10^{-12} \, \text{W/m}^2 \)).

When a sound level increases by 30 dB, the intensity increases by a factor of 1000 because \( 10^{3} = 1000 \). This shows the power of the logarithmic nature of the decibel scale, where large changes in intensity seem smaller and more manageable to analyze.

Pressure Amplitude

Pressure amplitude is a measure of the maximum change in pressure caused by a sound wave from its normal value. It relates to how loud a sound appears. While intensity looks at the energy of a sound wave, pressure amplitude is focused on how "forceful" or "strong" that wave feels.

There's a direct relationship between intensity and pressure amplitude, expressed mathematically as:

• \( I \propto P_{\text{amplitude}}^2 \)

This means that intensity is proportional to the square of the pressure amplitude. Therefore, when the intensity of a sound increases, the pressure amplitude also increases. In the exercise, when the intensity increases by a factor of 1000, the pressure amplitude increases by about \( \sqrt{1000} \), which is approximately \( 32 \). This relationship helps to understand how changes in intensity affect the perceived loudness of a sound.

Intensity Factor

An intensity factor is a dimensionless number that describes how the intensity of a sound has changed compared to another reference intensity. A change in sound intensity can indicate how much energy is being transferred by the wave over a specific area.

In the decibel formula:

• \( L_{dB} = 10 \log_{10} \left(\frac{I}{I_0}\right) \)

when the sound level changes by a certain decibel amount, we can calculate the intensity factor. For an increase of 30 dB, the intensity factor is:

• \( \frac{I}{I_0} = 10^{3} = 1000 \)

This indicates how much the energy carried by the sound wave has increased relative to its initial intensity. Understanding this factor can help in applications ranging from audio recording and sound engineering, to environmental noise control.