# Loudspeaker Question (Sound and Intensity)

• K_Physics
In summary, the intensity of sound at a distance of 3.0 m from a spherical loudspeaker emitting sound isotropically at 10 W into an anechoic chamber is 0.0884 W/m^2. The ratio of the wave amplitude at a distance of 4.0 m to that at a distance of 3.0 m can be found by calculating the surface areas at each distance and taking the ratio of their respective intensities. The relationship between intensity and amplitude may be proportional, but it is not needed to solve this problem.
K_Physics

## Homework Statement

Suppose a spherical loudspeaker emits sound isotropically at 10 W into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance d = 3.0 m from the center of the source? (b) What is the ratio of the wave amplitude at d = 4.0 m to that at d = 3.0 m?

## Homework Equations

[/B]
I = P/A

I ∝ A^2 (According to my TA)

## The Attempt at a Solution

[/B]
Part A:

I = P/A

I = 10/4π(3.0)^2 = 0.0884 W/m^2

Part B:

According to my TA, Intensity is proportional to the amplitude squared. I was wondering if I could use this relationship to answer part B.

Intensity at 3.0 m = 0.0884 W/m^2
Amplitude at 3.0 m = 0.297 m

Intensity at 4.0 m = 0.0497 W/m^2
Amplitude at 4.0 m = 0.223 m

Ratio: 0.223/0.297 = 0.751

Not sure if this is the correct approach or if the relationship I ∝ A^2 is correct or not.

Last edited by a moderator:
K_Physics said:

## Homework Statement

Suppose a spherical loudspeaker emits sound isotropically at 10 W into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance d = 3.0 m from the center of the source? (b) What is the ratio of the wave amplitude at d = 4.0 m to that at d = 3.0 m?

## Homework Equations

[/B]
I = P/A

I ∝ A^2 (According to my TA)

## The Attempt at a Solution

[/B]
Part A:

I = P/A

I = 10/4π(3.0)^2 = 0.0884 W/m^2

Part B:

According to my TA, Intensity is proportional to the amplitude squared. I was wondering if I could use this relationship to answer part B.

Intensity at 3.0 m = 0.0884 W/m^2
Amplitude at 3.0 m = 0.297 m

Intensity at 4.0 m = 0.0497 W/m^2
Amplitude at 4.0 m = 0.223 m

Ratio: 0.223/0.297 = 0.751

Not sure if this is the correct approach or if the relationship I ∝ A^2 is correct or not.
You are already given the power, so you only need to figure out what the surface areas are at the different radii to get the Intensity in W/m^2.

Intensity may be proportional to the amplitude of the sound pressure wave squared, but you don't need that in this problem.

## 1. What is a loudspeaker?

A loudspeaker is a device that converts electrical signals into sound waves. It consists of a diaphragm, or cone, which vibrates in response to the electrical signal and creates sound waves. The sound waves are then amplified and projected out of the speaker for us to hear.

## 2. How does a loudspeaker produce sound?

A loudspeaker produces sound by converting electrical energy into mechanical energy. When an electrical signal is sent through the speaker, it causes the diaphragm to vibrate at a specific frequency, which creates sound waves.

## 3. What is the difference between sound and intensity?

Sound refers to the physical sensation of hearing, while intensity is a measure of the amount of energy in a sound wave. Sound can be described as loud or soft, while intensity is measured in decibels and determines how loud a sound is.

## 4. How does the intensity of a sound affect our perception?

The intensity of a sound directly affects our perception of loudness. The higher the intensity, the louder the sound will be perceived. However, there are other factors such as frequency and distance that can also impact our perception of sound.

## 5. How can I measure the intensity of a sound?

The intensity of a sound can be measured using a sound level meter. This device measures the sound pressure level in decibels (dB). It can also be calculated using the formula I = P/A, where I is intensity, P is power, and A is the cross-sectional area of the sound wave.

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