- #1

Rorshach

- 136

- 0

This is a simple problem from a textbook I am reading, and everything below is written word by word and sign by sign from said textbook. Formulas given in the book just don't give the result authors claim they do:

An input of 1 W produces a SPL of 115 dB at 1 m. What is the SPL at 6.1 m?

SPL = 115 - 10log(0.22/1) = 115 - 15.7 = 99.3 dBThe assumption made in the 20log(6.1) factor is that the loudspeaker is operating in a free field and that inverse square law is valid in this case. This is a reasonable assumption for a 20-ft distance if the loudspeaker is remote from reflecting surfaces. A loudspeaker is rated at a sound- pressure level of 115 dB on axis at 1 m with 1 W into 8Ω. If the input were decreased from 1 to 0.22 W, what would be the sound pressure level at 1 m distance?

SPL = 115 - 10log(0.22/1)

= 115 - 6.6

= 108.4 dB

Note that 10log is used because two powers are being compared.

## Homework Statement

An input of 1 W produces a SPL of 115 dB at 1 m. What is the SPL at 6.1 m?

## Homework Equations

## The Attempt at a Solution

SPL = 115 - 10log(0.22/1) = 115 - 15.7 = 99.3 dBThe assumption made in the 20log(6.1) factor is that the loudspeaker is operating in a free field and that inverse square law is valid in this case. This is a reasonable assumption for a 20-ft distance if the loudspeaker is remote from reflecting surfaces. A loudspeaker is rated at a sound- pressure level of 115 dB on axis at 1 m with 1 W into 8Ω. If the input were decreased from 1 to 0.22 W, what would be the sound pressure level at 1 m distance?

SPL = 115 - 10log(0.22/1)

= 115 - 6.6

= 108.4 dB

Note that 10log is used because two powers are being compared.

Last edited by a moderator: