Question

You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?

Short answer

Move to 0.15 meters away for the sound level to be 60 dB.

Step-by-step solution

Convert Decibels to Intensity Ratio

The sound level in decibels (dB) is a logarithmic measure of the ratio of the intensity of a sound wave to a reference intensity. We can use the formula \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \), where \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity (\( 10^{-12} \text{ W/m}^2 \)). We first calculate the intensity ratio for both 20.0 dB and 60.0 dB.

Calculate Intensities at Both Distances

For 20 dB: \( 20 = 10 \cdot \log_{10} \left( \frac{I_1}{I_0} \right) \). Solving gives \( \frac{I_1}{I_0} = 10^2 = 100 \); hence, \( I_1 = 100I_0 \).For 60 dB: \( 60 = 10 \cdot \log_{10} \left( \frac{I_2}{I_0} \right) \). Solving gives \( \frac{I_2}{I_0} = 10^6 = 1,000,000 \); hence, \( I_2 = 1,000,000I_0 \).

Use Inverse Square Law

Sound intensity is inversely proportional to the square of the distance from the source: \( I \propto \frac{1}{r^2} \). Given \( I_1 = 100I_0 \) at 15.0 m, and we want \( I_2 = 1,000,000I_0 \), we use the relationship \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \).

Solve for the New Distance

Plug the values into the inverse square relationship: \( \frac{100I_0}{1,000,000I_0} = \left( \frac{r_2}{15} \right)^2 \). This simplifies to \( \frac{1}{10,000} = \left( \frac{r_2}{15} \right)^2 \). Solving for \( r_2 \), we get \( r_2 = 15 \times \frac{1}{100} = 0.15 \text{ meters} \).

Key concepts

Decibel Scale

Sound levels are measured using the decibel (dB) scale, which is a logarithmic way to describe a ratio. The decibel scale helps us understand how loud a sound is compared to a reference sound. The reference intensity, usually denoted as \( I_0 \), is set at the threshold of hearing, which is \( 10^{-12} \text{ W/m}^2 \). This threshold represents the quietest sound that a healthy human ear can detect.
To calculate the intensity level of a sound in decibels, we use the formula:

• \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \)

Here, \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity. This formula shows that every increase of 10 dB represents a tenfold increase in intensity. For example, a sound that is 60 dB is not just three times louder than a 20 dB sound, but rather 10,000 times more intense. This logarithmic nature is why smaller dB changes represent larger changes in intensity, which is crucial in understanding sound dynamics.

Inverse Square Law

The inverse square law is a fundamental principle that describes how sound intensity decreases with distance. According to this law, the intensity of sound is inversely proportional to the square of the distance from the sound source. Mathematically, we express this relationship as:

• \( I \propto \frac{1}{r^2} \)

Here, \( I \) denotes the intensity of the sound, while \( r \) represents the distance from the sound source. This means that when the distance is doubled, the sound intensity becomes one-fourth as intense. In the context of our exercise, moving from 15 meters to a closer distance significantly increases the sound intensity.To solve such problems, we typically set up a ratio using the intensities at two different distances. If \( I_1 \) is the intensity at the first distance \( r_1 \), and \( I_2 \) is the intensity at the desired distance \( r_2 \), then:

• \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \)

Solving this equation will help you find how close you need to move to hear a sound more clearly.

Sound Intensity

Sound intensity is a measure of the power carried by sound waves per unit area, usually expressed in watts per square meter (\( \text{W/m}^2 \)). It indicates how much sound energy passes through a particular area and is a key factor in determining how loud a sound is perceived by a listener.
In our example, we are given two sound levels: 20 dB and 60 dB. To find the sound intensity at these levels, we first use the decibel formula:

• At 20 dB: \( I_1 = 100 I_0 \)
• At 60 dB: \( I_2 = 1,000,000 I_0 \)

These calculations show the vast difference in sound intensity between the two levels. Understanding the sound intensity allows us to apply the inverse square law effectively. By knowing the sound intensity at a starting distance, you can find out how close you need to be to a sound source to experience a desired intensity. This essential concept highlights the relationship between intensity, distance, and perceived loudness, which all stem from the principles of physics governing sound.