Which harmonic frequencies can be detected in a tube closed at one end whose fundamental frequency is 200 Hz?
200 Hz
300 Hz
400 hz
600 Hz
1000 Hz
It's a select multiple and I said the last three which was wrong.
You were guessing. wavelength has to be a odd multiple of 200. If you dont get that, there are two of them
node at closed end, max at the open end.
Draw the picture.
first boom at 1/4 of wavelength = length of tube L
say the speed of sound is c
c = wavelength /period = wavelength* f
now we are told that 1/4 wavelength is L
c = 4 L*200 = 800 L
so the next time we have a node at the closed and a max at the open end is when
wavelength = (4/3)L
c = (4/3)L*f = 800 L
f = 800 *3/4 = 600 Hz <-------
the next time is when
5/4 wavelength = L
c = (4/5)L*f= 800 L
f = 1000 Hz <--------
200, 600, 1000
In a tube closed at one end, only odd harmonics are present. Therefore, the correct harmonic frequencies that can be detected in this tube with a fundamental frequency of 200 Hz are:
200 Hz (fundamental frequency)
600 Hz (3rd harmonic)
1000 Hz (5th harmonic)
To determine which harmonic frequencies can be detected in a tube closed at one end, we need to understand the concept of harmonics.
In a tube closed at one end, the fundamental frequency is the frequency of the first harmonic, which is also the lowest possible frequency that can be produced. In this case, the given fundamental frequency is 200 Hz.
The harmonic series for a tube closed at one end can be expressed as follows:
1st harmonic (fundamental frequency): 200 Hz
2nd harmonic: 3 times the fundamental frequency = 3 * 200 Hz = 600 Hz
3rd harmonic: 5 times the fundamental frequency = 5 * 200 Hz = 1000 Hz
4th harmonic: 7 times the fundamental frequency = 7 * 200 Hz = 1400 Hz
And so on.
Therefore, the correct options for the harmonic frequencies that can be detected in the given tube closed at one end are:
- 200 Hz (fundamental frequency)
- 600 Hz
- 1000 Hz