Question

The two nearest harmonics of a tube closed at one end and open at another end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

Answer 1

We know that a tube closed at one end and open at another end will form a closed organ pipe. So we will use the concept of frequency of closed organ pipe in this question. We will find the difference in successive frequencies of a closed organ pipe. This is equal to twice the fundamental frequency. We will use this relation and arrive at the correct answer.

Complete step by step answer:
We know that in this given question as the tube is open at one end and closed at one end so it will form a closed organ pipe. Now we know that frequency of a closed organ pipe is given by the formula
$f = \dfrac{{(2n - 1)u}}{{4l}}$
Where f, is the frequency
u is the velocity of sound wave
It is the length of pipe.
Now we know that frequency of nth vibration in closed organ pipe is given by ${f_n} = \dfrac{{(2n - 1)u}}{{4l}}$
Then the frequency of ${(n + 1)^{th}}$ harmonic vibration is given by ${V_{n + 1}} = \left[ {\dfrac{{2(n + 1) - 1}}{{4l}}} \right] = \dfrac{{(2n + 1)u}}{{4l}} = 260Hz$
Then $\dfrac{{(2n - 1)u}}{{4l}} = 220$
So ${V_{n + 1}} - {V_n} = 260 - 220$
That means $\left[ {\dfrac{{(2n + 1) - (2n - 1)u]}}{{4l}}} \right] = 40$
$\dfrac{{2u}}{{4l}} = 40$
Which translates to $\dfrac{u}{{2l}} = 40$
Therefore, fundamental frequency for $n = 0 \Rightarrow \left[ {\dfrac{{2(0) + 1]u}}{{4l}}} \right] = \dfrac{v}{{4l}} = \dfrac{{40}}{2} = 20$

Therefore, the fundamental frequency of the system is 20 Hz

Note:
One point to be Note:d here is the concept about the closed and open organ pipe. One should keep in mind the errors in formula of frequency for nth vibration. Also, the difference between frequencies between nth and $(n + 1)$ vibration is twice that of frequency.