The speed of sound in air is 350m/sec the frequency of the fundamental node emitted by tube length 50cm open one end is:
T = 0.50*4/340
do
f = 1/T = 340/2 = 170 Hz
http://www.physicsclassroom.com/class/sound/Lesson-5/Closed-End-Air-Columns
its one end is close so we will cosider it as closed or open i mean we have to apply the closed formula or open one
To find the frequency of the fundamental node emitted by a tube with one end open, given the speed of sound in air and the length of the tube, we need to use the formula for the fundamental frequency of a tube open at one end:
f = v / (2L)
where:
- f is the frequency of the sound wave,
- v is the speed of sound in air, and
- L is the length of the tube.
In this case, the speed of sound in air is given as 350 m/s and the length of the tube is 50 cm, which is equivalent to 0.5 m.
Plugging these values into the formula, we have:
f = 350 m/s / (2 * 0.5 m)
Simplifying this equation, we get:
f = 350 m/s / 1 m
Finally, solving for f:
f = 350 Hz
Therefore, the frequency of the fundamental mode emitted by a tube with one end open and a length of 50 cm is 350 Hz.