How is Wave Intensity Calculated?

[Janiceleong26]

Homework Statement

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..

 

[BvU]

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 )

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...

 

[Janiceleong26]

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 )

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?

 

[BvU]

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.

 

[Janiceleong26]

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??

 

[BvU]

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

 

[Janiceleong26]

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 a lot!