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A harmonic mode of a closed pipe of length 22 cm resonates when excited by a source frequency of 1875 Hz. Find the harmonic mode that resonates and the number of nodes present in it. (Velocity of sound in air is 330 m/s).
a) first harmonic, 1 node
b) third harmonic, 1 node
c) third harmonic, 2 nodes
d) fifth harmonic, 4 nodes
e) fifth harmonic, 3 nodes

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Last updated date: 22nd Jun 2024
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Answer
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Hint: A harmonic mode is said to resonate with a certain source frequency. This implies that the frequency of that particular harmonic node will be equal to the source frequency. Only odd harmonics will be present in a closed pipe.

Formula Used:
The natural frequencies of a closed pipe are given by, ${\nu _n} = \left( {n + \dfrac{1}{2}} \right)\dfrac{v}{{2l}}$ where $v$ is the velocity of sound in air, $n = 0{\text{, }}1{\text{, }}2{\text{,}}....$ is the number of the harmonics and $l$ is the length of the pipe.

Complete step by step answer:
Step 1: List the parameters that are known from the question.
The length of the closed pipe is $l = 22{\text{cm}}$ .
A particular harmonic resonates with an external source of frequency ${\nu _s} = 1875{\text{Hz}}$ .
The velocity of sound in air is $v = 330{\text{m/s}}$ .
Step 2: Find the harmonic that resonates with the external source.
The natural frequencies of a closed pipe are given by, ${\nu _n} = \left( {n + \dfrac{1}{2}} \right)\dfrac{v}{{2l}}$ -------- (1)
where $v$ is the velocity of sound in air, $n = 0{\text{, }}1{\text{, }}2{\text{,}}....$ is the number of the harmonics and $l$ is the length of the pipe.
The required harmonic can be found by calculating the natural frequency of each harmonic using equation (1).
Substituting for $n = 0$ in equation (1) we get the fundamental mode as ${\nu _0} = \dfrac{v}{{4l}}$ ------- (2)
Now, substitute the values for $v = 330{\text{m/s}}$ and $l = 22{\text{cm}}$ in the above relation.
Then the fundamental frequency is ${\nu _0} = \dfrac{{330}}{{4 \times 22 \times {{10}^{ - 2}}}} = 375{\text{Hz}}$
Now, ${\nu _0} \ne {\nu _s}$ and thus the first harmonic does not resonate with the source.
Substituting for $n = 1$ in equation (1) we get the third harmonic ${\nu _3} = \dfrac{{3v}}{{4l}}$
Now, substitute the values for $v = 330{\text{m/s}}$ and $l = 22{\text{cm}}$ in the above relation.
Then the third harmonic is ${\nu _3} = \dfrac{{3 \times 330}}{{4 \times 22 \times {{10}^{ - 2}}}} = 1125{\text{Hz}}$
Now, ${\nu _3} \ne {\nu _s}$ and thus the third harmonic does not resonate with the source.
Substituting for $n = 2$ in equation (1) we get the fifth harmonic ${\nu _5} = \dfrac{{5v}}{{4l}}$
Now, substitute the values for $v = 330{\text{m/s}}$ and $l = 22{\text{cm}}$ in the above relation.
Then the fifth harmonic is ${\nu _5} = \dfrac{{5 \times 330}}{{4 \times 22 \times {{10}^{ - 2}}}} = 1875{\text{Hz}}$
Now, ${\nu _5} = {\nu _s} = 1875{\text{Hz}}$
Thus the fifth harmonic resonates with the source.
The number of nodes present will be 3(including the one at the end).
Therefore the correct option is e.

Note: Alternate method
The third and fifth harmonics can also be expressed in terms of the first harmonic as $3{\nu _0}$ and $5{\nu _0}$ respectively.
Substituting the value for the fundamental frequency ${\nu _0} = 375{\text{Hz}}$ we get, third harmonic $3{\nu _0} = 3 \times 375 = 1125{\text{Hz}}$ and fifth harmonic $5{\nu _0} = 5 \times 375 = 1875{\text{Hz}}$
Also, when substituting the value for length in equation (1) a unit conversion from cm to m must be done.