Find Antinode Closest to x=0.25m in Standing Wave

[lha08]

Homework Statement

Consider the standing wave: 0.005sin(30x)cos(420t) m. Locate the antinode closest to x=0.25 m.

Homework Equations

The Attempt at a Solution

I'm not really sure how to tackle this problem except that i know that antinodes occur at the crests of the standing wave and that the distance between antinodes is lamda/2. In the answers it says that: kXn=(2n+1)pi/2... Xn=(2n+1)pi/(2k) and when you plug in n=2 it is 0.261 m...does anyone know how they got the initial formula?

 

[Doc Al]

Consider the spatial factor sin(kx). What values of x give the maximum amplitude?

 

[lha08]

Consider the spatial factor sin(kx). What values of x give the maximum amplitude?

aren't they pi/2, 3pi/2, 5pi/2, 7pi/2...etc?

 

[Doc Al]

aren't they pi/2, 3pi/2, 5pi/2, 7pi/2...etc?

Good. But that's the value for kx, not x. (In this problem, k = 30.) How would you write that series using n, where n = 0, 1, 2, and so on? Compare that to the formula given.