a rock group is playing in a club. Sound emerging outodoors from an open door spreads uniformly in all directions. If the decibel level is 70 dB at a distance of 1.0m from the door, at whiat distance is the music just barely audible to a person with a normal threshold of hearing. Disregard absorption
To determine at what distance the music is just barely audible, we need to understand the concept of the inverse square law. According to this law, the intensity of sound decreases with the square of the distance from the source.
In this case, we know that the decibel level is 70 dB at a distance of 1.0 meter from the door. We also know that the threshold of hearing (the minimum sound intensity detectable by a person with normal hearing) is typically around 0 dB.
The formula to calculate the sound intensity in decibels is:
I2 = I1 * (r1/r2)^2
Where:
I1 is the initial sound intensity at distance r1 (known)
I2 is the final sound intensity at distance r2 (to be determined)
Since we want to find the distance where the sound is just barely audible, we set I2 = 0 (minimum threshold of hearing).
Now let's set up the equation:
0 dB = 70 dB * (1.0 m / r2)^2
Since dB is a logarithmic scale, we need to convert the decibel values to intensity levels first. We can do this by using the formula:
I = 10^(dB/10)
Applying this formula to the above equation, we get:
1 = 10^(70/10) * (1.0 m / r2)^2
Now, we can solve for r2:
(1.0 m / r2)^2 = 1 / 10^(70/10)
Taking the square root of both sides:
1.0 m / r2 = sqrt(1 / 10^(70/10))
r2 = 1.0 m / sqrt(1 / 10^(70/10))
r2 ≈ 145.4 meters
Therefore, the music is just barely audible at a distance of approximately 145.4 meters from the open door.