
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) $2$
(B) $4$
(C) $7$
(D) $8$
Answer
580.8k+ views
Hint: Beats is given by ‘\[{{\text{f}}_{\text{o}}}{\text{- }}{{\text{f}}_{\text{c}}}\]’, where \[{{\text{f}}_{\text{o}}}\] is fundamental frequency of open organ pipe and \[{{\text{f}}_{\text{c}}}\] fundamental frequency of closed organ pipe.
Formula used:
$
{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 ${f_o}$
${f_c} = Lo\,(given)$
Where,
${L_c} = $length of closed organ pipe
$Lo = $open organ pipe
As we know,
${f_c} = \dfrac{v}{{4{L_c}}}$
${f_o} = \dfrac{v}{{4{L_o}}}$
Where\[{\text{V = }}\]velocity of sound.
Therefore,
${F_o} = 2{f_c}$ (since${{\text{L}}_o} = {L_c}$)……(i)
and
${f_o} - {f_c} = 2\,$ from question…..(ii)
From (i) and (ii)
$
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 $f_o^1 = \dfrac{v}{{2(\dfrac{{10}}{2})}} = 2{f_o} = 2 \times 4Hz = 8Hz$
When length of the closed pipe is doubled, its frequency of fundamentals is $f_c^1 = \dfrac{v}{{2(Lc)}} = \dfrac{1}{2}{f_c} = \dfrac{1}{2} \times 2Hz = 1Hz$
Therefore, number of beats produced per second $ = 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 \[7\].
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.
Formula used:
$
{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 ${f_o}$
${f_c} = Lo\,(given)$
Where,
${L_c} = $length of closed organ pipe
$Lo = $open organ pipe
As we know,
${f_c} = \dfrac{v}{{4{L_c}}}$
${f_o} = \dfrac{v}{{4{L_o}}}$
Where\[{\text{V = }}\]velocity of sound.
Therefore,
${F_o} = 2{f_c}$ (since${{\text{L}}_o} = {L_c}$)……(i)
and
${f_o} - {f_c} = 2\,$ from question…..(ii)
From (i) and (ii)
$
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 $f_o^1 = \dfrac{v}{{2(\dfrac{{10}}{2})}} = 2{f_o} = 2 \times 4Hz = 8Hz$
When length of the closed pipe is doubled, its frequency of fundamentals is $f_c^1 = \dfrac{v}{{2(Lc)}} = \dfrac{1}{2}{f_c} = \dfrac{1}{2} \times 2Hz = 1Hz$
Therefore, number of beats produced per second $ = 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 \[7\].
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.
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