Wave Intensity: Videos & Practice Problems

Wave intensity measures the energy a wave produces over a distance, defined as power divided by surface area. For three-dimensional waves, intensity decreases with distance due to the inverse square law, where I = P4πr2. As waves radiate outward, the energy spreads over a larger area, leading to lower intensity at greater distances. Understanding this relationship is crucial for analyzing sound and other wave phenomena.

Wave Intensity

In the study of wave mechanics, understanding wave intensity is crucial as it relates to the energy a wave transmits over a specific distance. Waves can be classified into one-dimensional, two-dimensional, and three-dimensional types based on their propagation. One-dimensional waves, such as those moving along a string, carry energy in a straight line from point A to point B. In contrast, two-dimensional waves, like ripples on a pond, spread outwards in a plane, while three-dimensional waves, such as sound waves from a loudspeaker, radiate energy in all directions.

Wave intensity is defined as the amount of energy per unit time (power) distributed over a surface area. The formula for wave intensity (I) can be expressed as:

I = \frac{P}{A}

where P is the power of the wave and A is the surface area over which the energy is spread. For three-dimensional waves, the surface area is represented by the formula for the surface area of a sphere, which is:

A = 4\pi r^2

Here, r is the distance from the source of the wave. Therefore, the intensity of a three-dimensional wave can be calculated using the equation:

I = \frac{P}{4\pi r^2}

The units for wave intensity are watts per meter squared (W/m²).

For example, if a loudspeaker produces 500 watts of power, and we want to find the wave intensity at a distance of 10 meters from the source, we can substitute the values into the intensity formula:

I = \frac{500}{4\pi (10)^2}

Calculating this gives:

I = \frac{500}{4\pi \cdot 100} = \frac{500}{400\pi} \approx 0.4 \text{ W/m}^2

This example illustrates how wave intensity decreases with distance due to the spreading of energy over a larger surface area. Understanding these concepts is essential for analyzing wave behavior in various physical contexts.

A sound source radiates sound waves in all directions. At a distance of 4 m from the source, you measure the wave intensity to be 0.06 W/m². How much sound energy does the source emit in 1 hour if the power output is constant?

• A. 724 J
• B. 10,860 J
• C. 1.07 J
• D. 43,400 J

Answer: (D)

Example 1

In this discussion, we explore the relationship between wave intensity and the average power output of the sun, utilizing real data about sunlight as an electromagnetic wave. The intensity of sunlight when it reaches Earth is approximately 1360 watts per square meter, and it takes about 500 seconds for sunlight to travel from the sun to Earth.

To find the average power output of the sun, we start with the wave intensity equation:

$$I = \frac{P_{\text{average}}}{A}$$

Here, \(I\) represents intensity, \(P_{\text{average}}\) is the average power output, and \(A\) is the surface area over which the power is distributed. Since the sun radiates light uniformly in all directions, the area can be modeled as the surface area of a sphere, given by:

$$A = 4\pi r^2$$

To proceed, we need to determine the distance \(r\) between the sun and Earth. This can be calculated using the relationship between speed, distance, and time:

$$v = \frac{\Delta x}{\Delta t}$$

In this case, the speed \(v\) is the speed of light, approximately \(c = 3 \times 10^8\) meters per second. Rearranging the equation gives us:

$$r = v \cdot \Delta t$$

Substituting the known values, we find:

$$r = (3 \times 10^8 \, \text{m/s}) \cdot (500 \, \text{s}) = 1.5 \times 10^{11} \, \text{m}$$

Now that we have the distance, we can substitute \(r\) back into the intensity equation to find the average power output of the sun:

$$P_{\text{average}} = I \cdot A = I \cdot 4\pi r^2$$

Plugging in the values:

$$P_{\text{average}} = 1360 \, \text{W/m}^2 \cdot 4\pi (1.5 \times 10^{11} \, \text{m})^2$$

After performing the calculations, we find that the average power output of the sun is approximately:

$$P_{\text{average}} \approx 3.85 \times 10^{26} \, \text{W}$$

This value closely aligns with the known power output of the sun, illustrating the immense energy it produces every second. This calculation not only highlights the application of physics in understanding celestial phenomena but also emphasizes the vast scale of energy involved in solar radiation.

The Inverse-Square Law for Intensity

Understanding wave intensity is crucial when analyzing how sound waves propagate from a source, such as a siren or loudspeaker. The intensity of a wave, denoted as \( I \), is defined as the power per unit area, and it can be affected by the distance from the source. The relationship between intensity and distance is governed by the inverse square law, which states that the intensity of a wave decreases as the distance from the source increases.

When a wave travels outward from a point source, the power \( P \) emitted remains constant, but the surface area over which this power is distributed increases. The surface area \( A \) of a sphere is given by the formula:

\[ A = 4\pi r^2 \]

where \( r \) is the radius (or distance from the source). As the distance \( r \) increases, the surface area increases, leading to a decrease in intensity. This can be expressed mathematically as:

\[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \]

To compare the intensity at two different distances, \( r_1 \) and \( r_2 \), we can set up the following relationship based on the inverse square law:

\[ I_1 \cdot 4\pi r_1^2 = I_2 \cdot 4\pi r_2^2 \]

By simplifying this equation, we arrive at:

\[ \frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 \]

This equation indicates that the ratio of the intensities is equal to the square of the ratio of the distances. It is important to note that the intensities \( I_1 \) and \( I_2 \) are inversely related to the distances \( r_1 \) and \( r_2 \).

For example, if we know the intensity \( I_1 \) at a distance \( r_1 \) and want to find the distance \( r_2 \) where the intensity \( I_2 \) is known, we can rearrange the equation to solve for \( r_2 \):

\[ r_2 = r_1 \sqrt{\frac{I_1}{I_2}} \]

In a practical scenario, if a siren produces an intensity \( I_1 = 0.25 \) at a distance \( r_1 = 15 \) meters, and we want to find the distance \( r_2 \) where the intensity drops to \( I_2 = 0.01 \), we can substitute these values into the rearranged equation:

\[ r_2 = 15 \sqrt{\frac{0.25}{0.01}} \]

Calculating this gives:

\[ r_2 = 15 \sqrt{25} = 15 \times 5 = 75 \text{ meters} \]

This result confirms that as the intensity decreases, the distance \( r_2 \) is indeed greater than \( r_1 \), illustrating the principle that intensity diminishes with increasing distance from the source. Understanding this relationship is essential for applications in acoustics and wave physics.

You measure the intensity from a sound source to be 0.3 W/m² at a distance of 3.4 m. What will the intensity be if you walk closer to the source, to a distance of 2.5 m?

• A. 0.41 W/m²
• B. 0.16 W/m²
• C. 0.55 W/m²
• D. 0.22 W/m²

Answer: (C)

Here’s what students ask on this topic:

What is wave intensity and how is it calculated?

Wave intensity measures the energy a wave produces over a certain distance. It is defined as the power (P) of the wave divided by the surface area (A) over which the energy is spread. For three-dimensional waves, the surface area is a sphere, so the formula for intensity (I) is:

$I=\frac{P}{4\pi r^2}$

where r is the distance from the source. The units for intensity are watts per meter squared (W/m²).

How does wave intensity change with distance?

Wave intensity decreases with distance due to the inverse square law. As waves radiate outward from a source, the energy spreads over a larger surface area. The intensity (I) at a distance (r) from the source is given by:

$I=\frac{P}{4\pi r^2}$

As the distance (r) increases, the surface area (4πr²) increases, causing the intensity to decrease. This relationship is crucial for understanding phenomena like sound waves and light intensity.

What is the inverse square law for wave intensity?

The inverse square law for wave intensity states that the intensity (I) of a wave is inversely proportional to the square of the distance (r) from the source. Mathematically, it is expressed as:

$I=\frac{P}{4\pi r^2}$

This means that if the distance from the source is doubled, the intensity is reduced to one-fourth. This law is essential for understanding how wave energy dissipates over distance.

How do you calculate the distance at which wave intensity falls to a certain value?

To calculate the distance at which wave intensity falls to a certain value, you can use the inverse square law. If you know the initial intensity (I₁) at a distance (r₁) and the desired intensity (I₂), you can find the new distance (r₂) using:

$\frac{I1}{I2}=\frac{r^2}{r^1}$

Rearrange to solve for r₂:

$r^2=\sqrt{r^1\times \frac{I}{1}\times \frac{1}{I}}$

This allows you to find the distance at which the intensity falls to the desired value.

What are the units of wave intensity?

The units of wave intensity are watts per meter squared (W/m²). This unit measures the power (in watts) distributed over a surface area (in square meters). Intensity quantifies how much energy a wave transmits through a given area per unit time.