Problem 47 - Pearson Physics

Question

If the power of a speaker is doubled and the area through which the sound is emitted is also doubled, does the intensity of the sound increase, decrease, or stay the same? Explain.

Step-by-Step Solution

Answer

The intensity of the sound stays the same.

Step 1: Understand the Formula for Intensity

The intensity of sound, denoted by \( I \), is given by the formula \( I = \frac{P}{A} \), where \( P \) is the power of the sound source, and \( A \) is the area through which the sound is emitted.

Step 2: Analyze the Initial Conditions

Initially, the power of the speaker is \( P \) and the area is \( A \). Thus, the initial intensity \( I_1 \) is \( I_1 = \frac{P}{A} \).

Step 3: Apply Changes to Power and Area

According to the problem, the power is doubled to \( 2P \), and the area is also doubled to \( 2A \).

Step 4: Calculate the New Intensity

Substitute the new values of power and area into the intensity formula: \( I_2 = \frac{2P}{2A} = \frac{P}{A} \).

Step 5: Compare Initial and New Intensity

Compare the initial intensity \( I_1 = \frac{P}{A} \) with the new intensity \( I_2 = \frac{P}{A} \). Since they are equal, the intensity remains the same.

Key Concepts

Power of Sound Source

The power of a sound source is an essential concept when discussing sound intensity. It represents the rate at which energy is emitted by the sound source over time. Think of it as the amount of energy that the sound source, such as a speaker, puts out continuously.
The power of a sound source is usually measured in watts (W). More power generally means more energy is being used to produce sound, which affects how loud or soft a sound can be. For instance, doubling the power of a speaker means it can emit twice the energy per unit of time.
When calculating the sound intensity, knowing the power is crucial.

• Higher power results in more intense sounds.
• Power and area together determine the intensity of emission.

In the problem, the speaker's power is doubled. This change tells us that the speaker's energy output is increased, but as we will see, intensity also depends on the area, making it vital to understand both aspects.

Area of Sound Emission

The area of sound emission relates to the surface space through which sound waves propagate. Imagine sound waves traveling from a speaker; they spread out in every direction, making an imaginary sphere around the source. This area determines how broadly the sound spreads out from its source.
When we talk about the area in the context of sound, it is crucial to consider how it interacts with power. The larger the area, the more the sound has to spread out, which affects its intensity.
In simple terms:

• If you increase the area, the sound has to travel over more space, which can reduce the intensity per unit area.
• Doubling the area through which sound is emitted, as in our problem, means the sound is trying to spread over double the surface space.

This balancing act between power and area is at the heart of determining the intensity of sound.

Intensity Formula

The sound intensity formula provides a clear mathematical relationship between power, area, and intensity. It is defined as the amount of sound power passing through a unit area.
The formula is given by:
\[ I = \frac{P}{A} \]
where

• \( I \) is the intensity,
• \( P \) is the power of the sound source, and
• \( A \) is the area through which the sound is emitted.

This formula indicates that sound intensity is directly proportional to power but inversely proportional to the area.
In our exercise, even though both power and area are increased twofold, the formula shows us that intensity doesn't change because:
\[ I_2 = \frac{2P}{2A} = \frac{P}{A} \]
This illustrates that if changes in power and area are equal, the intensity remains constant, providing key insight into sound distribution dynamics.