Final answer:
The number of watts required at a loudspeaker to achieve a sound pressure level of 70 dB SPL with 10 dB SPL of headroom at a distance of 14 feet is approximately 2 to 2.5 watts, after considering the inverse square law and loudspeaker sensitivity.
Step-by-step explanation:
To calculate the number of watts required at the loudspeaker with a given sound pressure level (SPL), sensitivity, headroom, and distance from the listener, we can use the inverse square law and the concept of loudspeaker sensitivity.
The loudspeaker has a sensitivity of 88 dB SPL for 1 watt of power at a distance of 1 meter. The sound pressure requirement is 70 dB SPL, but we also need to provide an additional 10 dB SPL of headroom, meaning the target SPL is 80 dB SPL at the listener's position.
First, we need to calculate the SPL loss over distance using the inverse square law. A doubling of the distance results in a loss of approximately 6 dB. At a distance of 14 feet (approximately 4.27 meters), the distance has doubled approximately 2.14 times from the reference distance (1 meter), resulting in a loss of about 12.84 dB (2.14 × 6 dB).
To achieve the target SPL of 80 dB at 14 feet, we need to compensate for the 12.84 dB loss due to distance. This requires us to set the speaker to produce 92.84 dB SPL at 1 meter. Since the speaker produces 88 dB SPL with 1 watt, we need an additional 4.84 dB gain.
Because every 10 dB represents a tenfold increase in power, a change of 3 dB represents approximately a doubling of power. Therefore, 4.84 dB is slightly more than a doubling, suggesting that approximately 2 to 2.5 watts would be required to achieve the desired headroom and SPL at the listener's position. However, these calculations can vary based on the specific properties of the room and speaker design.