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Question: A closed pipe and an open pipe of same length produce \[2\] beats, when they are set into vibrations...

A closed pipe and an open pipe of same length produce 22 beats, when they are set into vibrations simultaneously in their fundamental mode. If the length of the open pipe is halved, and that of closed pipe is doubled, and if they are vibrating in the fundamental mode, then the number of beats produced is:
A. 44
B. 77
C. 22
D. 88

Explanation

Solution

As we know that formula for frequency which depends on the velocity and length of the pipe. By considering two equations, one for closed pipe and another is for open pipe this question can be solved.

Formula Used:
fc=v4Lc{f_c} = \dfrac{v}{{4{L_c}}}
fo=v2Lo{f_o} = \dfrac{v}{{2{L_o}}}
Herefc{f_c} is the frequency of the closed pipe and fo{f_o} is the frequency of the open pipe and vv is the velocity and LO{L_O} is the length of open pipe andLc{L_c} is the length of closed pipe.

Complete step by step answer:
Consider for a closed pipe, the frequency of fundamental mode is –
fc=v4Lc{f_c} = \dfrac{v}{{4{L_c}}}
Where, fc{f_c} is frequency of closed pipe, vvis velocity of sound in air and Lc{L_c}is length of closed pipe.
Consider for an open pipe, the frequency of fundamental mode is
fo=v2Lo{f_o} = \dfrac{v}{{2{L_o}}}
Where, fo{f_o}is frequency of open pipe, vvis velocity andLo{L_o}is the length of open pipe.
Given that, Lc{L_c}= Lo{L_o}
Also, fo=2fc{f_o} = 2{f_c} ...... (A)
According to given statement
fofc=2{f_o} - {f_c} = 2 ……. (B)
Solving these equations, we get
fo=4Hz{f_o} = 4Hz
fc=2Hz{f_c} = 2Hz
As given in the statement when the length of the open pipe is halved then its frequency is
fo1=v2(Lo2)f_o^1 = \dfrac{v}{{2\left( {\dfrac{{{L_o}}}{2}} \right)}} =2fo=2×4Hz=8Hz2{f_o} = 2 \times 4Hz = 8Hz
As given in the statement when the length of closed pipe is doubled then its frequency is
fc1=v4(2Lc)f_c^1 = \dfrac{v}{{4\left( {2{L_c}} \right)}}
fc1\Rightarrow f_c^1 = 12fc\dfrac{1}{2}{f_c}
fc1\Rightarrow f_c^1 = 12×2Hz\dfrac{1}{2} \times 2Hz
fc1\therefore f_c^1 =1Hz1Hz
Then, number of beats produced is =fo1fc1=81=7f_o^1 - f_c^1 = 8 - 1 = 7

So, option B is correct.

Additional information:
Air Column can be found in various musical instruments where it is enclosed with a hollow metal tube. To conserve space it is coiled upon itself several times. It is almost nearer to one meter in length. Suppose if the end of the tube is left uncovered and thus allowing the sound waves to reach it then that end is termed as an open end. Various instruments operate on the mechanism of the open-end air column that is when the end of the tube is completely uncovered. Suppose if one of the ends of the tube is left uncovered in the surrounding atmosphere and another end of the tube is covered then it is termed as a closed air column.

Note: Fundamental note is the note of the lowest frequency of the periodic waveform. Above the fundamental notes, are called overtones. Cell phones that we use in our daily life are based on radio frequencies.Some pipe organs and their columns are composed within the orchestra. It is possible to convert open tube air columns to close tube air columns.