What frequencies will a 1.55 m long tube produce in the audible range (20 Hz - 2

Question

What frequencies will a 1.55 m long tube produce in the audible range (20 Hz - 20,000 Hz) at 15.0°C for the following cases?

(a) the tube is closed at one end lowest frequency 54.8 Correct: Your answer is correct. Hz

second lowest frequency 164.5 Correct: Your answer is correct.

Hz highest frequency (rounded to the nearest Hz) 274.19 Incorrect: Your answer is incorrect. Hz

(b) the tube is open at both ends

lowest frequency 109.6 Correct: Your answer is correct.

Hz second lowest frequency 219.2 Correct: Your answer is correct.

Hz highest frequency (rounded to the nearest Hz) 19999.8 Incorrect: Your answer is incorrect. Hz

Explanation / Answer

For a closed tube oscillation will occur when 2 * L = 1/2 lambda
That gives the first frequency.
It will also occur at 3/2 5/2 7/2 ......(2n-1)/2
Or put simply IF you are correct with your lowest frequency then the remaining frequencies are
1,3,5,7,9 ...... (2n-1) * f0

There is no maximum. Just a maximum that we can hear.
so for the information we have been given

(2n-1) * f0 < 20000 you can solve this for the highest value of n and for the highest frequency.
But as there are so many frequencies that are so close together the answer is meaningless in any practical situation.
In practice these are the musical effects of playing a clarinet. This instrument is almost unique in using this 1,3,5 series for the notes.

All other instruments are designed with either open both ends ( trumpet, flute, recorder etc)
or closed both ends ( violin, piano )

For the second series.
If it is open at both ends there is a path difference on reflection of 1/2 wavelength.
Giving NO net path difference from the reflection alone.

So we get 2L = n lambda where n is any whole number.

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