A pipe 0.14 meters long and capped at one end exists in a room where the speed of sound is 350 m/s. Determine the three lowest frequencies that will resonate within the tube.

[Enter the frequencies in order from lowest to highest.]

http://www.physicsclassroom.com/class/sound/Lesson-5/Closed-End-Air-Columns

remember
period T = wavelength/speed
and
f = 1/T = 350/wavelength

here the lowest frequency is when .14 meter = half a wavelength
so wavelength = .28 meter
lowest f = 350/.28

To determine the three lowest frequencies that will resonate within the tube, we need to use the formula for the resonant frequencies of closed-end pipes.

The formula for the resonant frequencies of a closed-end pipe is:

f = (n * v) / (4 * L)

Where:
- f is the frequency
- n is the harmonic number (1, 2, 3, etc.)
- v is the speed of sound
- L is the length of the pipe

Given:
- Length of the pipe (L) = 0.14 meters
- Speed of sound (v) = 350 m/s

We need to find the three lowest frequencies, so we can start with n = 1, 2, and 3. Let's calculate:

For n = 1:
f1 = (1 * 350) / (4 * 0.14) = 625 Hz

For n = 2:
f2 = (2 * 350) / (4 * 0.14) = 1250 Hz

For n = 3:
f3 = (3 * 350) / (4 * 0.14) ≈ 1875 Hz

Therefore, the three lowest frequencies that will resonate within the tube are: 625 Hz, 1250 Hz, and 1875 Hz.