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## GR0877 Problem 47

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) For a pipe of length $L$, closed at one end and open at the other the resonant frequencies are given by $\displaystyle f = \frac{v}{\lambda} = \frac{n v}{4L}$, where $n = 1, 3, 5, \dots$ and $v$ is the speed of the sound. Thus, a fundamental frequency is $\displaystyle f(n = 1) = \frac{v}{4L}$ and the next harmonic has a frequency $\displaystyle f(n = 3) = \frac{3 v}{4L}$. For this next harmonic $f(n = 3) = 3f(n = 1) = 3 \cdot 131 = 393$ (Hz).

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