Answer to Question #155551 in Physics for Olivia

Question

You have a closed pipe which resonates it’s first harmonic when the pipe is .23m long.  What is the wavelength and frequency of the tuning form at 343m/s?

Answer 1

$\frac{\lambda}{2}=0.23\to \lambda=0.46(m)$ . Answer

$\nu=\frac{v}{\lambda}=\frac{343}{0.46}=745.6(Hz)$ . Answer