Animated Solution for Physics - Waves: A sound absorber attenuates the sound level by 20 db. The intensity decreases by a factor of

Human Perception

Imagine you are standing near a roaring jet engine, and then you walk into a perfectly quiet library. The physical energy—the intensity—of the sound waves hitting your eardrums changes by a factor of over a trillion! If we tried to measure sound using raw intensity (Watts per square meter), the numbers would be incredibly clumsy to work with.

To solve this, physicists use the decibel (dB) scale. It is a logarithmic scale that compresses this massive range of physical intensities into a neat, human-friendly scale from $0$ to $120$ dB. The fundamental equation bridging the physical world to our perception is:

Here, $I$ is the actual intensity of the sound, and $I_0$ is the reference intensity (usually $10^{-12}{\text{ W/m}}^2$), which represents the absolute quietest sound a healthy human ear can detect.

In our problem, a sound wave passes through an absorber, and its sound level drops by $20\text{ dB}$. Let's define our states. Before hitting the absorber, the sound has an intensity $I_1$ and a sound level ${\beta }_1$. After passing through, it has a reduced intensity $I_2$ and a sound level ${\beta }_2$.

The problem tells us that the attenuation (the drop in sound level) is exactly $20\text{ dB}$. Mathematically, we write this as:

Now, let's substitute our logarithmic definitions into this simple difference equation:

At first glance, this might look messy, but logarithms have a beautiful property that cleans things up instantly. Recall the quotient rule for logarithms: $\log A-\log B=\log (A/B)$.

When we apply this rule, the reference intensity $I_0$ in the denominators perfectly cancels out! We are left with a pure ratio of the two intensities:

To find out exactly how much the physical intensity dropped, we just need to solve for the ratio $\frac{I_1}{I_2}$. First, divide both sides by $10$:

Now, we convert this logarithmic equation back into its exponential form. Since the base of our logarithm is $10$, we get:

This means $I_1=100\times I_2$, or conversely, $I_2=\frac{I_1}{100}$.

The physical intensity of the sound didn't just drop by a little bit; it plummeted to one-hundredth of its original value! This perfectly illustrates the magic of the decibel scale: a seemingly small drop of $20\text{ dB}$ corresponds to a massive $100$-fold decrease in actual physical energy.