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Question: A closed organ pipe and an organ pipe of same length produce \(2\dfrac{{beats}}{{\sec ond}}\) when t...

A closed organ pipe and an organ pipe of same length produce 2beatssecond2\dfrac{{beats}}{{\sec ond}} when they are set into vibrations together in fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced will be
(A)7(A)7
(B)4(B)4
(C)8(C)8
(D)2(D)2

Explanation

Solution

This question will be solved by using the formula for frequency which depends on the velocity and length of the pipe. On considering two equations, one for closed pipe and the other one from the open pipe this question can be easily solved.

Complete step by step solution:
For a closed pipe, the frequency of fundamental mode is calculated by,
fc=v4Lc{f_c} = \dfrac{v}{{4{L_c}}}
Where,
fc{f_c} is known as the frequency of the closed pipe
vv is known as the velocity of sound in air
Lc{L_c} is known as the length of the closed pipe
Now let us 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 known as the frequency of the open pipe
vv is known as the velocity of sound in air
Lo{L_o} is known as the length of the open pipe
In this question, we are given that,
Lc=Lo{L_c} = {L_o}
Also, fo=2fc....(1){f_o} = 2{f_c}....(1)
From the given statement, we can also say that,
fofc=2....(2){f_o} - {f_c} = 2....(2)
On solving equation (1) and equation (2), we get,
fo=4Hz{f_o} = 4Hz
fc=2Hz{f_c} = 2Hz
As it is clearly given in the question that when the length of the open pipe is cut to half its value, then,
fo1=v2(Lo2)=2fof_o^1 = \dfrac{v}{{2\left( {\dfrac{{{L_o}}}{2}} \right)}} = 2{f_o}
fo1=2×4=8Hzf_o^1 = 2 \times 4 = 8Hz
As it is clearly given in the question that when the length of the closed pipe is doubled its value, then,
fc1=v4(2Lc)f_c^1 = \dfrac{v}{{4\left( {2{L_c}} \right)}}
fc1=12fcf_c^1 = \dfrac{1}{2}{f_c}
fc1=12×2f_c^1 = \dfrac{1}{2} \times 2
fc1=1Hzf_c^1 = 1Hz
Thus, the number of beats which are produced are,
fo1fc1=81f_o^1 - f_c^1 = 8 - 1
fo1fc1=7f_o^1 - f_c^1 = 7
So, the correct answer is (A)7(A)7.

Note:
Fundamental note is defined as the note of the lowest frequency which is present in the periodic waveform. The notes above the fundamental notes are called overtones. Cell phones which are used by us in our daily life are based on the principle of radio frequencies. Few of the pipe organs and their columns are made within the orchestra. It is very much possible to convert an open tube air column to a closed tube air column.