
It we study the vibration of a pipe open at both ends, then the following statement is not true:
(A) Open end will be antinode
(B) Odd harmonics of the fundamental frequency will be generated
(C) All harmonics of the fundamental frequency will be generated
(D) Pressure change will be maximum at both ends
Answer
147.9k+ views
Hint A pipe open at both ends is struck which causes vibration throughout. Study vibration wave characteristics and determine whether there is an antinode at the ends and whether harmonics of fundamental frequency will be generated. Conclude your answer using it.
Complete Step By Step Solution
A pipe of length L is struck, which causes it to vibrate. Now, the vibration will be along its length L and won’t surpass the limit. Hence, we can say that the open ends will be anti-nodes, where the displacement is maximum or minimum in a wave system.
Now, let us find out whether the harmonics of fundamental frequency will be generated or not. Now harmonics is said to be a set of frequencies that are multiples of fundamental frequency. Fundamental frequency is defined as the lowest frequency produced by the instrument or body. For a tube with open ends the fundamental frequency is given as
\[{f_0} = \dfrac{{n\upsilon }}{{2L}}\], where \[\upsilon \]the frequency of vibrating waves, L is is the length of the tube and n is the number of harmonics.
Now , n can hold all positive integer values, thus can generate both an odd number of fundamental harmonics and even number of fundamental harmonics, thus ruling out option (b) and (c).
We found out that the opening ends of the pipe are antinodes. Thus the opening ends of the pipe will be exposed to only atmospheric pressure. Due to the presence of anti-node at the ends, there won’t be much pressure difference from atmospheric pressure to the pressure inside the pipe, rather the pressure difference will be higher at the middle of the pipe. It is given the exact opposite way in the options section.
Thus, option(d) is the right answer for the given question.
Note The fundamental frequency is defined as the lowest possible frequency produced by any vibrating instrument. In harmonics terms, it is called the first harmonic of any instrument. First harmonic consists of 2 nodes and one antinode.
Complete Step By Step Solution
A pipe of length L is struck, which causes it to vibrate. Now, the vibration will be along its length L and won’t surpass the limit. Hence, we can say that the open ends will be anti-nodes, where the displacement is maximum or minimum in a wave system.
Now, let us find out whether the harmonics of fundamental frequency will be generated or not. Now harmonics is said to be a set of frequencies that are multiples of fundamental frequency. Fundamental frequency is defined as the lowest frequency produced by the instrument or body. For a tube with open ends the fundamental frequency is given as
\[{f_0} = \dfrac{{n\upsilon }}{{2L}}\], where \[\upsilon \]the frequency of vibrating waves, L is is the length of the tube and n is the number of harmonics.
Now , n can hold all positive integer values, thus can generate both an odd number of fundamental harmonics and even number of fundamental harmonics, thus ruling out option (b) and (c).
We found out that the opening ends of the pipe are antinodes. Thus the opening ends of the pipe will be exposed to only atmospheric pressure. Due to the presence of anti-node at the ends, there won’t be much pressure difference from atmospheric pressure to the pressure inside the pipe, rather the pressure difference will be higher at the middle of the pipe. It is given the exact opposite way in the options section.
Thus, option(d) is the right answer for the given question.
Note The fundamental frequency is defined as the lowest possible frequency produced by any vibrating instrument. In harmonics terms, it is called the first harmonic of any instrument. First harmonic consists of 2 nodes and one antinode.
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