- #1

- 3

- 3

- Homework Statement
- Ans: C

I tried to use the intensity formulae and after manipulating it got an expression which gave me wrong answer.

It seems the areas are not equivalent somehow, any explanation?

- Relevant Equations
- I = P/A

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter iQadmat
- Start date

- #1

- 3

- 3

- Homework Statement
- Ans: C

I tried to use the intensity formulae and after manipulating it got an expression which gave me wrong answer.

It seems the areas are not equivalent somehow, any explanation?

- Relevant Equations
- I = P/A

- #2

- #3

- 3

- 3

Hi and thanks for welcoming me!Hi @iQadmat and welcome to PF.

How does the intensity depend on the amplitude? You can read the amplitudes of the two waves off the graph and compare.

I know the intensity is directly proportional to amplitude^2 frequency^2 and inversely proportional to radius^2, but my point was the areas (which I showed in ss) are different somehow and I want to understand the differences between them and what they represent.

So far my understanding*

[The initial 'A' is showing the area at which the power from source is acting upon but the final 'A' (later derived from 'F') is not the initial 'A' but is the area of the wave particle which times by pressure of wave equals 'F' .]

- #4

- 13,190

- 6,488

- #5

- 3

- 3

I got it now, Thanks for your help!

Share:

- Replies
- 8

- Views
- 608

- Replies
- 3

- Views
- 1K

- Replies
- 24

- Views
- 2K

- Replies
- 1

- Views
- 963

- Replies
- 5

- Views
- 630

- Replies
- 11

- Views
- 712

- Replies
- 13

- Views
- 792

- Replies
- 4

- Views
- 796

- Replies
- 7

- Views
- 808

- Replies
- 8

- Views
- 388