Principle Determining the Loudness (Intensity) of Sound

AI Thread Summary
The discussion centers on how the loudness of sound is affected by distance, particularly when a radio is moved three times further away. It highlights that loudness is determined by the energy of sound waves, which spreads out over a larger area as distance increases, leading to a decrease in perceived loudness. The conversation also touches on the misconception that amplitude remains constant with distance, clarifying that amplitude decreases as sound waves propagate spherically. Participants explore the relationship between energy, amplitude, and distance, emphasizing that the relevant factor is the surface area of the sphere rather than its volume. Overall, the dialogue enhances understanding of sound wave propagation and energy distribution.
Impulse
Messages
19
Reaction score
0
A test question I was asked earlier in the year has been hanging around in the back of my mind:

"If a radio is moved 3x further away from you, how is its loudness affected?"

To answer this question one needs to know what determines the loudness of sound. What determines the loudness of sound?

Is the loudness of sound determined solely by the amplitude of the sound wave? I answered based on this reasoning, and said that the radio would be equally loud 3x further away because its sound waves would have equal amplitude at distances x and 3x.
 
Physics news on Phys.org
Impulse said:
A test question I was asked earlier in the year has been hanging around in the back of my mind:

"If a radio is moved 3x further away from you, how is its loudness affected?"

To answer this question one needs to know what determines the loudness of sound. What determines the loudness of sound?

Is the loudness of sound determined solely by the amplitude of the sound wave? I answered based on this reasoning, and said that the radio would be equally loud 3x further away because its sound waves would have equal amplitude at distances x and 3x.
The "loudness" ("amplitude") depends on energy. Energy is conserved but the sound wave spreads out in a sphere. So at 3 times the distance the same amount of energy is spread out over 9 times the area.

(Do you really believe that a radio sounds just as loud no matter how far away you are?)
 
Loudness is determined by energy. Got it. Based on that, it makes intuitive sense that the further away one is from the source, the less energy one will receive (i.e. heat energy from the sun decreases with distance).

Still, I don't fully understand how the energy is spreading out into space. When I imagine a wave, I think of a two-dimensional structure like a sine wave or the longitudinal wave animation here: http://www.acs.psu.edu/drussell/demos/waves/wavemotion.html

In the above animation, because the longitudinal wave is in a rectangular pool and generated from a wall of the pool, and not a point source, the amplitude and energy are constant irrespective of distance travelled.

If I imagine light (or sound) as particles, it is easy to picture the energy spreading out into a sphere. I don't see that picture with waves, because I picture a constant amplitude and don't see it diminishing with distance (i.e. I don't image a graph of sine with decaying amplitude).

Intuitively I don't think it's crazy for a radio to be just as loud 3x further away. Isn't a laser just as intense 3x further away from its source?

Also, if energy travels outward like a sphere, why is the relationship not inverse cube with distance? The formula for volume of a sphere = (4/3)pi(r^3).

How do you picture the propagation and energy distribution of sound waves? Do you picture waves, or particles / packets of energy?
 
Impulse said:
When I imagine a wave, If I imagine light (or sound) as particles, it is easy to picture the energy spreading out into a sphere. I don't see that picture with waves, because I picture a constant amplitude and don't see it diminishing with distance (i.e. I don't image a graph of sine with decaying amplitude).

The amplitude of a spherical wave does indeed decrease with distance from the center. For a two-dimensional analog, have you ever dropped a stone into a pool of water and watched the circular ripples expand? Their amplitude also decreases.

Isn't a laser just as intense 3x further away from its source?

A laser beam does expand gradually with distance, but the description is more complicated than with a simple point source.

Also, if energy travels outward like a sphere, why is the relationship not inverse cube with distance? The formula for volume of a sphere = (4/3)pi(r^3).

What matters is the surface area of the sphere (4πr2), not its volume. Your ears intercept part of this surface area as it reaches your location a distance r from the source.
 
  • Like
Likes 1 person
jtbell said:
The amplitude of a spherical wave does indeed decrease with distance from the center. For a two-dimensional analog, have you ever dropped a stone into a pool of water and watched the circular ripples expand? Their amplitude also decreases.

I was imagining that as I wrote my reply. Is there a way to calculate the rate at which the amplitudes decrease?

I read that tsunamis undergo little energy loss with distance travelled. Are they not a point-source wave? Also, the amplitude of waves increases as it nears the shore, "shoaling", yet its total energy couldn't be increasing. Is there more to energy of a wave than just amplitude?

In general, what determines the energy in a wave?

jtbell said:
A laser beam does expand gradually with distance, but the description is more complicated than with a simple point source.

Okay, cool.

jtbell said:
What matters is the surface area of the sphere (4πr2), not its volume. Your ears intercept part of this surface area as it reaches your location a distance r from the source.

Makes sense. The energy doesn't fill the entire space at once (volume) but expands out through it (surface area). Thanks.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

Similar threads

Back
Top