Sound intensity (loudness) varies inversely as the square of the distance from the source. If a rock band has a sound intensity of 100 decibels 25 feet away from the amplifier, find the sound intensity 50 feet away from the amplifier.
twice the distance means 1/4 the loudness
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To solve this problem, we need to use the inverse square law for sound intensity. According to this law, the sound intensity (loudness) is inversely proportional to the square of the distance from the source.
Let's define the variables:
I1 = Initial sound intensity (loudness)
I2 = Final sound intensity (loudness)
d1 = Initial distance from the source
d2 = Final distance from the source
According to the problem, the initial sound intensity is 100 decibels when the initial distance is 25 feet from the amplifier. We need to find the sound intensity (I2) when the distance is 50 feet from the amplifier.
We can set up the proportional relationship as follows:
I1 / I2 = (d2^2) / (d1^2)
Plugging in the given values:
100 / I2 = (50^2) / (25^2)
Simplifying the equation:
100 / I2 = 2500 / 625
Now, we can cross-multiply and solve for I2:
100 * 625 = I2 * 2500
62,500 = I2 * 2500
To isolate I2, divide both sides of the equation by 2500:
I2 = 62,500 / 2500
I2 = 25
Therefore, the sound intensity 50 feet away from the amplifier is 25 decibels.
Using the inverse square law, we found that the sound intensity is reduced by a factor of 4 when the distance is doubled. In this case, the sound intensity is halved when the distance is doubled (from 25 to 50 feet), resulting in a decrease from 100 decibels to 25 decibels.