When SPL is doubled, by how much does the dB increase?

When SPL is doubled, by how much does the dB increase?

• A. 3 dB
• B. 5 dB
• C. 6 dB
• D. 10 dB

Answer: (C)

The correct answer is rooted in understanding the relationship between sound pressure level (SPL), measured in decibels (dB), and sound intensity. The decibel scale is logarithmic, meaning that each increase of 10 dB represents a tenfold increase in intensity. When the sound pressure level (SPL) is doubled, it does not simply double the intensity; rather, it leads to an increase in dB based on the logarithmic scale. Specifically, doubling the SPL results in an increase of approximately 6 dB. This is derived from the formula used to calculate decibels, where a change in intensity is expressed as: \[ \text{dB} = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \] In the case of doubling the SPL, the ratio \( \frac{I_2}{I_1} \) equals 2, and when you plug this into the logarithmic equation, you calculate: \[ \text{dB increase} = 10 \log_{10}(2) \approx 3 \text{ dB} \] However, since SPL is also influenced by a reference level

The correct answer is rooted in understanding the relationship between sound pressure level (SPL), measured in decibels (dB), and sound intensity. The decibel scale is logarithmic, meaning that each increase of 10 dB represents a tenfold increase in intensity.

When the sound pressure level (SPL) is doubled, it does not simply double the intensity; rather, it leads to an increase in dB based on the logarithmic scale. Specifically, doubling the SPL results in an increase of approximately 6 dB. This is derived from the formula used to calculate decibels, where a change in intensity is expressed as:

[

\text{dB} = 10 \log_{10} \left( \frac{I_2}{I_1} \right)

]

In the case of doubling the SPL, the ratio ( \frac{I_2}{I_1} ) equals 2, and when you plug this into the logarithmic equation, you calculate:

[

\text{dB increase} = 10 \log_{10}(2) \approx 3 \text{ dB}

]

However, since SPL is also influenced by a reference level