nign. windows, door, 1.2 m x 2 m. wide. Each surface has an Absorption coefficient of 1,000 Hz as shown in the table below Surface Wall Floor Ceiling Window Door Absorption coefficient 0.05 0.10 0.06 0.02 0.04 In this room, there is a loudspeaker located in the middle of the room with a sound power level is 60.0 dB at 1,000 Hz. Find the reverberation time, Room reverberation level (Le) and total

Principles of Physics: A Calculus-Based Text
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Chapter24: Electromagnetic Waves
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A room 10 m. long, 5 m. wide and 3.5 m. high. The wall has 2 windows, 1.5 m x 1 m wide. and 1
door, 1.2 m x 2 m. wide. Each surface has an Absorption coefficient of 1,000 Hz as shown in the
table below
Surface
Wall
Floor
Ceiling
Window
Door
Absorption coefficient
0.05
0.10
0.06
0.02
0.04
In this room, there is a loudspeaker located in the middle of the room with a sound power level
is 60.0 dB at 1,000 Hz. Find the reverberation time, Room reverberation level (Lev) and total
sound level received by the listener at a distance of 3 m from the loudspeaker (Let Q =1)
Transcribed Image Text:A room 10 m. long, 5 m. wide and 3.5 m. high. The wall has 2 windows, 1.5 m x 1 m wide. and 1 door, 1.2 m x 2 m. wide. Each surface has an Absorption coefficient of 1,000 Hz as shown in the table below Surface Wall Floor Ceiling Window Door Absorption coefficient 0.05 0.10 0.06 0.02 0.04 In this room, there is a loudspeaker located in the middle of the room with a sound power level is 60.0 dB at 1,000 Hz. Find the reverberation time, Room reverberation level (Lev) and total sound level received by the listener at a distance of 3 m from the loudspeaker (Let Q =1)
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