Animated Solution for Physics - Waves: A small speaker delivers $2\text{ W}$ of audio output. At what distance from the speaker will one detect $120\text{ dB}$ intensity sound? [Take, reference intensity of sound as $10^{-12}{\text{ W/m}}^2$]

Imagine you are standing at a certain distance from a small speaker. The speaker is pumping out sound energy at a constant rate, specifically with a power output of $P=2\text{ W}$. As the sound travels outward, it spreads over a larger and larger area, which means the intensity of the sound decreases the further away you get.

We are tasked with finding the exact distance, $r$, where the sound level is perceived as $120\text{ dB}$. This is a very loud sound, right at the threshold of pain for human hearing! To solve this, we need to bridge the gap between the perceived loudness (in decibels) and the physical intensity of the sound wave (in watts per square meter).

Loudness to Intensity

The relationship between the loudness $\beta$ in decibels and the physical intensity $I$ is given by the logarithmic formula:

Here, $I_0$ is the reference intensity, which represents the faintest sound a typical human ear can detect. It is a standard constant value: $I_0=10^{-12}{\text{ W/m}}^2$. We know our target loudness is $\beta =120\text{ dB}$. Let's substitute these values into our master equation:

Now, we need to carefully solve for $I$. First, divide both sides by 10:

To remove the logarithm, we take the antilog (base 10) of both sides. This means we raise 10 to the power of each side:

Multiplying both sides by $10^{-12}$ isolates $I$:

So, at the location where the sound is $120\text{ dB}$, the physical intensity of the sound wave is exactly $1{\text{ W/m}}^2$.

Intensity to Distance

Now that we have the intensity, how do we find the distance? We must remember that a small speaker acts like a point source, emitting sound waves uniformly in all directions. These waves form expanding spheres. The intensity $I$ at a distance $r$ is simply the total power $P$ divided by the surface area of the sphere at that radius:

We know $I=1{\text{ W/m}}^2$ and $P=2\text{ W}$. Let's substitute these into the intensity formula:

Now, we just need to rearrange this equation to solve for $r^2$:

Taking the square root of both sides gives us the distance $r$:

Using the approximation $\pi \approx 3.14159$, we get $\sqrt{2\pi }\approx \sqrt{6.283}\approx 2.506$. Therefore:

Since the options are given in centimeters, we multiply by 100 to convert meters to centimeters:

Rounding to the nearest integer, we get $40\text{ cm}$. This matches option (a) perfectly!