Question

The two nearest harmonics of a tube closed at one end and open at the other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?
(A) $ 40 Hz $
(B) $ 10 Hz $
(C) $ 20 Hz $
(D) $ 30 Hz $

Answer 1

Hint
According to the question there is a tube closed at one end. The nearest harmonics of the tube are 220Hz, 260Hz. A harmonic of such a wave is a wave with a frequency that is a positive integer multiple of the frequency of the original wave, known as the fundamental frequency. The original wave is also called the 1st harmonic, the following harmonics are known as higher harmonics.

Complete step by step answer
We know there is a tube which closed at one end and open at other end
So, we need to find the $ {n^{th}} $ harmonic of the tube
$\Rightarrow \therefore {\vartheta _n} = \dfrac{{(2n + 1)}}{2}(\dfrac{v}{{2L}}) $
$\Rightarrow {\vartheta _n} = \dfrac{{(2n + 1)v}}{{4L}}$ … [According to the formula, equation(1)]
Now, there is two nearest harmonic 220Hz and 260Hz,
Let, one is for $ {n^{th}} $ harmonic and one is for the nearest one which is say, $ {(n + 1)^{th}} $ harmonic.
Now, frequency for the $ {(n + 1)^{th}} $ harmonic is,
$\Rightarrow {\vartheta _{(n + 1)}} = \dfrac{{(2(n + 1) + 1)}}{{4L}}(v) = (\dfrac{{2n + 3}}{{4L}})(v) $
$ \therefore {\vartheta _{(n + 1)}} = (\dfrac{{2n + 3}}{{4L}})(v) $ … equation 2
We know the value of both frequency
Now, $ {\vartheta _{(n + 1)}} - {\vartheta _n} = 260 - 220 $ =40
Now put the value from equation 1 and equation 2
We get,
$\Rightarrow (\dfrac{{2n + 3}}{{4L}}).v - (\dfrac{{2n + 1}}{{4L}}).v = 40$
$\Rightarrow \dfrac{{2v}}{{4L}} = 40 $
$\Rightarrow \dfrac{v}{{2L}} = 40 $ … equation 3
For value of “n” we let the hole numbers like n=0,1,2,3,4........so on
Now, fundamental frequency for n=0
$\Rightarrow {\vartheta _0} = \dfrac{{(2(0) + 1)}}{{4L}}.v = (\dfrac{v}{{4L}}) $
$\therefore {\vartheta _0} = \dfrac{v}{{4L}} $
$\Rightarrow {\vartheta _0} = \dfrac{v}{{2L}}.\dfrac{1}{2}$
We know the value of $ \dfrac{v}{{2L}} $ which is 40
So, $ {\vartheta _0} = \dfrac{{40}}{2} = 20 $ Hz.
Option (C) is correct.

Note
The fundamental frequency is just the lowest possible frequency among all the natural frequencies of vibration of an object. For a pendulum/tuning forks, the fundamental frequency is the