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Question

One closed organ pipe has a length of 2.40 m. a. What is the frequency of the note played by this pipe? b. When a second pipe is played at the same time, a 1.40-Hz beat note is heard. By how much is the second pipe too long?

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 3Information given in the text are:

$L = 2.40\, \mathrm{m}$

$\textit{a.}$, $f = ?$

Frequency is given by:

$f = \dfrac{v}{\lambda}$

Therefore, we need to determine wavelength since the speed of sound is known $v = 343\, \mathrm{m/s}$:

$\begin{align*} \lambda &= 4L\\ &= 4 \cdot 4.20\, \mathrm{m}\\ &= 9.60\, \mathrm{m}\\ \end{align*}$

Frequency will, finally, be:

$f = \dfrac{343\, \mathrm{m/s}}{9.60\, \mathrm{m}}$

$\boxed{f = 35/7\, \mathrm{Hz}}$

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