# Fundamental Frequency of Two Pipe Organs

• Northbysouth
In summary, the problem involves two organ pipes, one of which is lengthened by 2.50 cm. The missing part asks to find the frequency of the beat produced when playing together in their fundamental. The solution involves using the equations λ = nL/4, fn = nv/4L, and v = λF to find the fundamental frequencies of the two pipes, and then using the equation fBeat = fa-fb to find the frequency of the beat.
Northbysouth

## Homework Statement

Two organ pipes, open at one end but closed at the other, are each 1.18 m long. One is now lengthened by 2.50 cm

λ = nL/4

fn = nv/4L

v = λF

## The Attempt at a Solution

Here's what I tried

First I tried finding the fundamental frequency when their lengths were equal

f = (1)(343 m/s)/4(1.18m)
f = 72.66949153 Hz

I'm assuming that v = 343 m/s. It does not say that this is the case in the problem.
Then I tried finding the frequency of the pipe with the extension

fextended = (1)(343 m/s)/4(1.205m)
fextended = 71.16182573 Hz

Having found these two frequencies I then took of the average of them which gave me 71.916 Hz. Unsurprisingly this didn't work. Any suggestions?

Pipe need not be resonating in the fundamental mode. So take lambda =(2n +1)L/4 and proceed.

If you still need help for this problem, try using this equation

fBeat = fa-fb

Solve for frequency using f=(nv)/(4L) where fa is the fundamental frequency for the pipe at its original length and fb is the fundamental frequency for the pipe when it is extended.

And v=344m/s (speed of sound in air)

Sorry it's been so long since I've replied, its been a busy week.
But yes you're right

f_beat = f_a - f_b

So I found that if I take f_a to be

f_a = (1)(343 m/s)/4(1.18m)
f_a = 72.66949153 Hz

Then the pipe with the increased length

f_b = (1)(343 m/s)/4(1.205 m)
f_b = 71.16182573 Hz

Then
f_beat = 72.66949153 Hz - 71.16182573 Hz
f_beat = 1.507 Hz

Rounded to 3 sig figs, 1.51 Hz is the correct answer.

The problem statement is not complete. What is required in the problem?

You're right, it is missing a part; I don't know how I managed that. Sorry to waste your time. The missing part is:

a) Find the frequency of the beat they produce when playing together in their fundamental.

## What is the fundamental frequency of two pipe organs?

The fundamental frequency of two pipe organs is the lowest frequency produced by the two organs when played simultaneously. It is determined by the length, diameter, and material of the pipes, as well as the air pressure and temperature.

## How is the fundamental frequency of two pipe organs calculated?

The fundamental frequency can be calculated using the formula f=nv/2L, where f is the frequency, n is the number of nodes (1 for the fundamental frequency), v is the speed of sound, and L is the length of the pipe.

## Can the fundamental frequency of two pipe organs be adjusted?

Yes, the fundamental frequency can be adjusted by changing the length, diameter, or material of the pipes, as well as adjusting the air pressure and temperature. This can be done manually or electronically using a pipe organ tuner.

## Why is the fundamental frequency important in pipe organs?

The fundamental frequency is important because it determines the overall pitch of the pipe organ. It also serves as a reference point for tuning and harmonizing the other pipes in the organ.

## Do all pipe organs have the same fundamental frequency?

No, the fundamental frequency of pipe organs can vary depending on the design and construction of the organ. In addition, different types of pipe organs, such as baroque and romantic organs, may have different fundamental frequencies due to their distinct styles and historical periods.

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