Question

(a) Show that the intensity \(I\) is the product of the energy density \(u\) (energy per unit volume) and the speed of propagation \(v\) of a wave disturbance; that is, show that \(I=u v .(b)\) Calculate the energy density in a sound wave \(4.82 \mathrm{~km}\) from a \(47.5-\mathrm{kW}\) siren, assuming the waves to be spherical, the propagation isotropic with no atmospheric absorption, and the speed of sound to be \(343 \mathrm{~m} / \mathrm{s}\).

Short answer

The intensity (I) of a wave disturbance can indeed be expressed as the product of the energy density (u) and the speed of propagation (v). The energy density of the given sound wave is approximately \(0.0003 J/m^3\).

Step-by-step solution

Proving the Intensity Formula

The intensity of a wave can be formally defined as the power per unit area carried by a wave. Mathematically, we can express it like: \( I = \frac{P}{A} \). Power is characterized as the amount of energy used or produced in a certain amount of time. Therefore, it can be represented as \( P = \frac{E}{t} \). Energy density (u) is defined as energy per unit volume. Hence, \( E = uV \). Substituting this into the power equation, we get \( P = uV\frac{1}{t} \). Now, Velocity (v) of a wave is given by the distance travelled over time which implies that \( V = vt \). Again substituting this into the power equation, we arrive at \( P = uvt \). Finally, substituting this power equation into the intensity equation: \( I = \frac{uvt}{A} \). Considering \( vt = A \) (since the wave covers area A in time t), we replace it in the intensity equation to obtain, \( I = uv \). Hence, we've proved the relation that Intensity = Energy Density x Speed of Propagation.

Calculation of Energy Density in the Sound Wave

The power of the siren is given as 47.5 kW (which equals \(4.75 \times 10^4\) W because 1 kW = 1000 W) and the speed of sound is given as 343 m/s. To calculate the energy density, we should first calculate the intensity. Using the intensity formula derived earlier, \( I = \frac{P}{A} \), and taking into account that the sound wave is spherical and propagates isotropically, the area \( A \) is the surface area of a sphere, which equals \( 4\pi r^2 \). Here, \( r \) is the radius of the sphere (distance from the siren), which is given as 4.82 km or \( 4.82 \times 10^3 \) m (since 1 km = 1000 m). So, \( A = 4\pi (4.82 \times 10^3)^2 \) m². Substituting the values, we get the intensity \( I = \frac{4.75 \times 10^4}{4\pi (4.82 \times 10^3)^2} \) W/m². Now, from the formula derived in the last step, we know that \( I = uv \). Hence we can solve for u by rearranging this equation as \( u = \frac{I}{v} \). Substituting the known values \( I \) and \( v \) into this equation, we get \( u = \frac{4.75 \times 10^4}{4\pi (4.82 \times 10^3)^2 \times 343} \) J/m³. This is the energy density of the sound wave.

Key concepts

Understanding Energy Density

Energy density is a fundamental concept when studying waves and their behavior. It's defined as the amount of energy that is stored in a given system or region of space per unit volume. For sound waves, and waves in general, energy density can help us understand how much sound energy is present in a specific volume of air.

To visualize, imagine a balloon that's filling up with air. The energy density would be how much energy is inside the balloon divided by the volume of the balloon. In the context of sound waves produced by, for instance, a siren, the energy density tells us how much acoustic energy is packed into the space through which the sound is traveling. Energy density isn't static; it changes as the wave spreads out, typically diminishing with distance as the energy is distributed over a larger volume. This reduction is especially true for spherical waves, where the energy spreads out in all directions equally.

Wave Propagation Speed Explained

Wave propagation speed is the rate at which a wave travels through a medium, and it is a key player in determining a wave's characteristics. For sound waves, this speed is the speed of sound, which varies in different materials but is about 343 meters per second (m/s) in air at room temperature.

Why does wave speed matter? It's a big deal because it allows us to calculate how energy, like sound from a siren, travels over time and distance. It's similar to how you'd measure how fast a car is traveling. If the car goes faster, it'll cover more distance in less time. Similarly, a sound wave with a high speed will travel further in less time compared to one with a lower speed. Understanding this speed is crucial for numerous applications, including designing buildings for good acoustics or measuring distances using sonar.

The Nature of Spherical Waves

A spherical wave is a type of wave that propagates outwards in all directions from a source in a spherical manner. This description fits many natural phenomena such as the ripples from a pebble dropped in water, or the sound from a loudspeaker in an open field.

Imagine throwing a stone into a still pond. The waves spread out in a circle from that point; if we could see this in three dimensions, it would be a sphere expanding outward. That's a spherical wave. The significant property of spherical waves is that as they spread out, their intensity decreases with the square of the distance from the source - this is known as the inverse square law. It's like how a flashlight's beam gets wider and dimmer as you shine it further away. This principle affects not just the light from a flashlight but also acoustics, which is why sound from a siren becomes fainter as you move away from it.

Determining the Power of a Wave

The power of a wave is the amount of energy it transfers per unit of time. For sound waves, power would be the total acoustic energy the siren emits every second in the form of sound. It's measured in watts (W), just like electrical power. This concept is vital: the higher the power of a sound source, the louder it can be heard over greater distances.

Power ties back to energy density and wave propagation speed. If we know the power of a source like a siren and how fast the sound is traveling, we can figure out how much sound energy is traveling through a particular area. This can relate to how loud the sound is or how much energy hits a surface, which is crucial for applications such as designing soundproofing in buildings or measuring the loudness of machinery.