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Question: A closed organ pipe and an open organ pipe of same length produce 2 beats per sec while vibrating in...

A closed organ pipe and an open organ pipe of same length produce 2 beats per sec while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of the closed organ pipe is doubled. Then the number of beats produced per second while vibrating in the fundamental mode is
(A) 22
(B) 44
(C) 77
(D) 88

Explanation

Solution

Beats is given by ‘fofc{{\text{f}}_{\text{o}}}{\text{- }}{{\text{f}}_{\text{c}}}’, where fo{{\text{f}}_{\text{o}}} is fundamental frequency of open organ pipe and fc{{\text{f}}_{\text{c}}} fundamental frequency of closed organ pipe.

Formula used:
fc=uL1L2 fc=v2Lo   {f_c} = \dfrac{u}{{{L_1}{L_2}}} \\\ {f_c} = \dfrac{v}{{2Lo}} \\\ \\\

Complete step by step solution:
Given data
Fundamental frequency for closed organ pipe fc and fundamental frequency for open organ pipe is fo{f_o}
fc=Lo(given){f_c} = Lo\,(given)
Where,
Lc={L_c} = length of closed organ pipe
Lo=Lo = open organ pipe

As we know,
fc=v4Lc{f_c} = \dfrac{v}{{4{L_c}}}
fo=v4Lo{f_o} = \dfrac{v}{{4{L_o}}}
WhereV = {\text{V = }}velocity of sound.
Therefore,
Fo=2fc{F_o} = 2{f_c} (sinceLo=Lc{{\text{L}}_o} = {L_c})……(i)
and
fofc=2{f_o} - {f_c} = 2\, from question…..(ii)
From (i) and (ii)
2fcfc=2 fc=2Hzandfo=4Hz  2{f_c} - {f_c} = 2 \\\ {f_c} = 2Hz\,and\,{f_o} = 4Hz \\\

When the length of the open pipe is halved its frequency of fundamentals is fo1=v2(102)=2fo=2×4Hz=8Hzf_o^1 = \dfrac{v}{{2(\dfrac{{10}}{2})}} = 2{f_o} = 2 \times 4Hz = 8Hz

When length of the closed pipe is doubled, its frequency of fundamentals is fc1=v2(Lc)=12fc=12×2Hz=1Hzf_c^1 = \dfrac{v}{{2(Lc)}} = \dfrac{1}{2}{f_c} = \dfrac{1}{2} \times 2Hz = 1Hz

Therefore, number of beats produced per second =fo1fc1=81=7= f_o^1 - f_c^1 = 8-1=7
So, the number of beats produced per second is while vibrating in the fundamental mode is 77.

Hence, (C) is correct.

Additional information: An organ pipe is a sound producing element of the pipe organ that resonates at a specific pitch when pressurized air is blown through it.

Note: In this type of question formula should be applied directly with all consistent in S.I. unit. The calculations are performed step by step in order to avoid mistakes and the various fundamentals used in organ pipe concept is followed thoroughly.