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?
λ2=0.23→λ=0.46(m)\frac{\lambda}{2}=0.23\to \lambda=0.46(m)2λ=0.23→λ=0.46(m) . Answer
ν=vλ=3430.46=745.6(Hz)\nu=\frac{v}{\lambda}=\frac{343}{0.46}=745.6(Hz)ν=λv=0.46343=745.6(Hz) . Answer
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