## University Physics with Modern Physics (14th Edition)

a) $$L = 0.328 m$$ b) $$λ = 1.312 m$$ c) $$f = 262.2 Hz$$
a) The formula for the frequency in a open pipe is: $f=\frac{nv}{2L}$ The question specifies fundamental frequency so n=1 (i.e. the first harmonic). $Length = \frac{nv}{2f}$ $Length = \frac{1\times344 m/s}{2\times524 Hz}$ $$Length = 0.328 m$$ b) In a closed pipe, the formula for the frequency is $f=\frac{nv}{4L}$. Since $f=\frac{v}{λ}$, we can rearrange for the wavelength and substitute the equation for the frequency to get: $λ = \frac{4L}{n}$ Therefore, $$λ = \frac{4\times\times\times0.328 m}{1} = 1.312 m$$ c) $f=\frac{nv}{4L}$. $$f = \frac{1\times344 m/s}{4\times0.328 m} = 262.2 Hz$$