1. The problem statement, all variables and given/known data What is the wavelength of the lowest note that can resonate within an air column 42 cm in length and closed at both ends. 2. Relevant equations Given: l = 42 cm n = 1 Required: [tex]\lambda[/tex] Analysis: l = [tex]\frac{n\lambda}{2}[/tex] therefore [tex]\lambda[/tex] = [tex]\frac{2l}{n}[/tex] I don't know how this came to be, could someone please explain. I've just gone back to high school after 15 years and would appreciate if you could also forward me on to some good links for rules for mathematical equations such as these. Thanks Scott
The steps to go from: [tex]l = \frac{n\lambda}{2}[/tex] to [tex]\lambda = \frac{2l}{n}[/tex] ? start at: [tex]l = \frac{n\lambda}{2}[/tex] first multiply both sides by 2. that gives: [tex]2l = 2\times \frac{n\lambda}{2}[/tex] on the right side, the 2 in the numerator cancels with the 2 in the denominator, so [tex]2l = n\lambda[/tex] Then divide both sides by n. [tex]\frac{2l}{n} = \frac{n\lambda}{n}[/tex] on the right side, the n in the numerator cancels with the n in the denominaotr. so, [tex]\frac{2l}{n} = \lambda[/tex] Then just switch sides. [tex]\lambda = \frac{2l}{n}[/tex]
No prob. Sounds like you're looking for a refresher on algebra. This seems to be a complete algebra course online: http://www.themathpage.com/alg/algebra.htm sosmath has quite a few examples: http://www.sosmath.com/algebra/solve/solve0/solve0.html Here's another page where you can practice: http://www.coolmath.com/algebra/algebra-practice-solving.html Hope this helps.