A multichoice question on intensity

[Charismaztex]

Homework Statement

A plane wave of amplitude A is incident on a surface of area S placed so that it is perpendicular to the direction of travel of the wave. The energy per unit time reaching the surface is E.
The amplitude of the wave is increased to 2 A and the area of the surface is reduced to
0.5 S.

How much energy per unit time reaches this smaller surface?
A) 4E
B) 2E
C) E
D) 0.5E

Homework Equations

Answer is B) by the way.

The Attempt at a Solution

I know that intensity is proportional to (aplitude)^2, but right now I'm a little confused.Thanks in advance,
Charismaztex

 

[Redbelly98]

The Attempt at a Solution

I know that intensity is proportional to (aplitude)^2, but right now I'm a little confused.

That's right. So as a first step in thinking about this, what would happen if all they did was to double the amplitude?

 

[Charismaztex]

so the energy would be proportional to (2A)^2=4A^2. The energy would quadruple. I also know that intensity is inversely proportional to the area, so if the area decreases by half, the intensity would increase by a factor of 2. Wouldn't that mean that the energy would increase by a factor of 8?

I think I'm missing something crucial here.

 

[Redbelly98]

so the energy would be proportional to (2A)^2=4A^2. The energy would quadruple.

You're essentially correct, but a better way to think of it is that intensity is proportional to amplitude^2.

Next hint: Energy per unit time = Intensity x Area.

 

[Charismaztex]

Wait, I think I may have got it. The amplitude doubles so that the intensity quadruples. But intensity= power/time =energy/(time x area S) so E=I x t x S. Hence if when amplitude doubles and surface area halves, E= 4I x t x 0.5S =2I x t x S Hence energy doubles.

 

[Redbelly98]

Yup, you got it.

 

[Charismaztex]

Thanks :)