Energy per unit time of a wave

In summary, intensity is related to amplitude as it is defined as the amount of energy that passes through unit area perpendicular to the wave direction in unit time. Amplitude, on the other hand, is the amount of extension with respect to equilibrium and can be measured in various units such as length, angle, or pressure. The relationship between intensity and amplitude can be represented by the equation I ∝ A^2, where I is intensity and A is amplitude. Additionally, intensity is equal to power per unit area, which can be expressed by the equation I = P/A, where P is power and A is area.
  • #1
Janiceleong26
276
4

Homework Statement


image.jpg

Is intensity related to this question? The answer is B.

Homework Equations


I^2 ∝ A^2
I=P/A,
where I is intensity, P is power, and A is area

The Attempt at a Solution


I tried using I=P/A, and after substituting, I get, A^2=E/S, which gives E=A^2 S
And then, the new wave now is (2A)^2= x/(1/2S), where x is the energy per unit time after the changes, and this gives x=8 A^2 S, which..does not give an answer..:oldmad:
 
Physics news on Phys.org
  • #2
Check your first relevant equation. (A is the amplitude, as in the exercise text). Is it right ? Later on you do something else (which seems better to me :smile: )

I = E/S where I is intensity, E is power and S is area (as in the exercise text -- see comment below )

You can't re-use the letter A for Area in the second equation.
Don't redefine given names with names that are in use already. It's confusing, makes your substitutions illegible, etc.
(I understand where this comes from, but doing an exercise you have to adapt a little... and in this case convert (substitute) to I = E/S even before writing down)


According to your equations, you have ## E \propto A^2 S## which is good and gives the proper answer.

Comment:
I must say that the exercise writer wrong-footed me too by using the letter E for energy per unit time. Very unusual, to say something decent...
 
  • #3
BvU said:
Check your first relevant equation. (A is the amplitude, as in the exercise text). Is it right ? Later on you do something else (which seems better to me :smile: )

I = E/S where I is intensity, E is power and S is area (as in the exercise text -- see comment below )

You can't re-use the letter A for Area in the second equation.
Don't redefine given names with names that are in use already. It's confusing, makes your substitutions illegible, etc.
(I understand where this comes from, but doing an exercise you have to adapt a little... and in this case convert (substitute) to I = E/S even before writing down)According to your equations, you have ## E \propto A^2 S## which is good and gives the proper answer.

Comment:
I must say that the exercise writer wrong-footed me too by using the letter E for energy per unit time. Very unusual, to say something decent...

Oh oh, sorry, yeah.. Was unconscious about it the A.
Ok thanks.. Umm, may I know how is intensity related to amplitude? And why is intensity equals to power per unit area? How does it come about?
 
  • #4
That's the definition...

You wrote ##I^2 \propto A^2## as relevant equation, but substituted ##I\propto A^2## in the attempt at solution...

The simplistic way I memorize it is to think of a weight hanging from a spring. The energy is ½ k ^x2 so proportional to amplitude2.
 
  • #5
BvU said:
That's the definition...

You wrote ##I^2 \propto A^2## as relevant equation, but substituted ##I\propto A^2## in the attempt at solution...

The simplistic way I memorize it is to think of a weight hanging from a spring. The energy is ½ k ^x2 so proportional to amplitude2.
Oh woops. Sorry again..
Oh I see, so intensity is sort of like the energy and amplitude is the extension??
 
  • #6
The intensity of a wave is defined as the amount of energy that passes through unit area perpendicular to the wave direction in unit time.
So it's energy per unit time per unit area.

Amplitude is amount of extension with respect to equilibrium (deviation from equilibrium). Can be length, but can also be angle, or pressure, or perhaps a few more things
 
  • #7
BvU said:
So it's energy per unit time per unit area.

Amplitude is amount of extension with respect to equilibrium (deviation from equilibrium). Can be length, but can also be angle, or pressure, or perhaps a few more things
I see, thanks alot!
 

1. What is energy per unit time of a wave?

The energy per unit time of a wave, also known as power, is the rate at which energy is transferred by a wave. It is measured in watts (W).

2. How is energy per unit time of a wave calculated?

The energy per unit time of a wave can be calculated by multiplying the amplitude of the wave by the frequency, and then squaring the result. This value is then divided by the characteristic impedance of the medium the wave is traveling through.

3. What is the relationship between energy per unit time and amplitude of a wave?

The energy per unit time of a wave is directly proportional to the square of the amplitude of the wave. This means that as the amplitude increases, the energy per unit time also increases.

4. How does energy per unit time of a wave affect the intensity of the wave?

The intensity of a wave is directly proportional to the energy per unit time. This means that as the energy per unit time increases, the intensity of the wave also increases.

5. How does the medium affect the energy per unit time of a wave?

The medium through which a wave travels can affect the energy per unit time. Different mediums have different characteristic impedances, which can alter the amount of energy that is transferred by the wave. Additionally, factors such as temperature and density of the medium can also affect the energy per unit time of a wave.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
556
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
785
  • Introductory Physics Homework Help
Replies
7
Views
216
  • Introductory Physics Homework Help
Replies
1
Views
808
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
770
Back
Top