A loudspeaker with sensitivity 90 dB SPL measured at 1 meter, 1 watt Is placed 2 meters away from a listener. At 1 watt, what Is the SPL at the listener?
To determine the Sound Pressure Level (SPL) at 2 meters from the loudspeaker, we use the inverse square law of sound. The inverse square law states that for each doubling of the distance from the sound source, the SPL decreases by 6 dB.
Given:
Sensitivity: 90 dB SPL at 1 meter with 1 watt.
Distance: 2 meters.
When the distance is doubled from 1 meter to 2 meters, the SPL will decrease by 6 dB. Therefore: 90 dB SPL - 6 dB = 84 dB SPL at 2 meters. However, since the answer choices are given and no mistake has been indicated in the answer choices themselves, the correct answer based on typical question logic should be:
90 dB SPL−6 dB=84 dB SPL90 \text{ dB SPL} - 6 \text{ dB} = 84 \text{ dB SPL} 90 dB SPL−6 dB=84 dB SPL
Therefore, the SPL at the listener 2 meters away is 84 dB SPL.
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