Question

A resonance tube of length \[1m\] is resonated with tuning with a tuning fork of frequency \[300Hz\]. If the velocity of sound in air is \[\dfrac{{300m}}{{{s^{ - 1}}}}\] then the number of harmonies produced in the tube will be.
(A) \[1\]
(B) \[2\]
(C) \[3\]
(D) \[4\]

Answer 1

Hint: Harmonies: Harmonies are the generated term used to describe the distinction of a sinusoidal wave formed by waveforms of different frequencies.
Resonance Harmonies: Resonance Harmonies is a multi potential pattern formation principle, that is the same mechanism that generates the fundamental resonance, can also create a range of patterns declined harmonies by the same essential principle.

Complete step by step answer:
We have to from the formula,
$
  {L_1} = \dfrac{\lambda }{4} = \dfrac{1}{4} = 0.25m, \\
  {L_2} = \dfrac{{3\lambda }}{4} = \dfrac{1}{4} = 0.75m, \\
 $
We get,

$\lambda = \dfrac{v}{f}\;{\text{ = }}\dfrac{{300}}{{300}}{\text{ = 1m}}{\text{.}}$

Let the lengths of resonant columns be\[{L_1}\], \[{L_2}\] and \[{L_3}\] therefore
For first resonance,

${L_1} = \dfrac{\lambda }{4} = \dfrac{1}{4} = 0.25m,$
For second resonance,

${L_2} = \dfrac{{3\lambda }}{4} = \dfrac{1}{4} = 0.75m,$
For third resonance.

${L_3} = \dfrac{{5\lambda }}{4} = \dfrac{1}{5} = 1.25m,$

It is clear that only two harmonies are possible because \[{L_3}\] is greater than the total length \[1m\] of the tube.

So, the correct answer is “Option B”.

Note:
The frequencies of the various harmonies are multiple of the frequency of the first harmonic; each harmonic frequency (\[{f_n}\]) is given by the equation \[{f_n} = n{f_1}\].