A fifth harmonic standing wave is set up in a closed-ended pipe that is 1.4 m long. If the frequency of the wave is 239 Hz, what is its velocity?

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A fifth harmonic standing wave is set up in a closed-ended pipe that is 1.4 m long. If the frequency of the wave is 239 Hz, what is its velocity?

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Answer 1 · 2018-03-21T05:34:34

$v = 268 \frac{m}{s}$ $v=268m/s$

Explanation:

It's important to remember our two formulas for open-ended pipes and closed-ended pipes when it comes to finding frequency in terms of harmonics:

Open-ended pipes: $f = (1) (\frac{v e l o c i t y}{2 L})$ $color(blue)(f=(1)((velocity)/(2L)))$
Closed-ended pipes: $f = (1) (\frac{v e l o c i t y}{4 L})$ $color(blue)(f=(1)((velocity)/(4L)))$

Obviously, we'll disregard the first formula for this problem, but it's important to remember both. That being said, let's plug in what we're given:

$239 Hz = (5) (\frac{v e l o c i t y}{4 (1.4 m)})$ $239"Hz" = (5)((velocity)/(4(1.4"m")))$

$47.8 Hz = (\frac{v e l o c i t y}{4 (1.4 m)})$ $47.8"Hz" = ((velocity)/(4(1.4"m")))$ $\frac{239 Hz}{5} = 47.8 Hz$ $color(blue)((239"Hz")/(5)=47.8"Hz"$

$47.8 Hz = (\frac{v e l o c i t y}{5.6 m})$ $47.8"Hz"=((velocity)/(5.6"m"))$ $4 \cdot 1.4 m = 5.6 m$ $color(blue)(4*1.4"m"=5.6"m")$

$v e l o c i t y = 267.68 \frac{m}{s}$ $velocity=267.68m/s$ $47.8Hz \cdot 5.6 m = 267.68 \frac{m}{s}$ $color(blue)(" 47.8Hz"*5.6"m"=267.68m/s)$

$v e l o c i t y = 268 \frac{m}{s}$ $velocity=268m/s$ $Assuming 3 significant figures$ $color(blue)("Assuming 3 significant figures")$

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A fifth harmonic standing wave is set up in a closed-ended pipe that is 1.4 m long. If the frequency of the wave is 239 Hz, what is its velocity?

1 Answer

Explanation:

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