Courses
Courses for Kids
Free study material
Offline Centres
More

A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio
A. $1:2$
B. $2:3$
C. $3:4$
D. $4:5$

seo-qna
Last updated date: 22nd Feb 2024
Total views: 358.8k
Views today: 3.58k
IVSAT 2024
Answer
VerifiedVerified
358.8k+ views
Hint: A ratio is a quantitative representation used to determine how much value of one thing is contained in the other thing. Here, we will use the formula used to calculate the frequency of the first overtone in an open pipe and closed pipe to calculate the ratio of lengths of pipe.

Complete step by step answer:
An overtone is a frequency generated by a substance or instrument which is generally greater than the fundamental frequency of a sound.
Let, the length of the open pipe is ${l_0}$ and the length of the closed pipe is ${l_c}$.
Now, the frequency of the first tone of an open pipe is given by
${\nu _o} = \dfrac{{2v}}{{2{l_o}}}$
Here, ${\nu _o}$ is the frequency of the first tone of the open pipe, $v$ is the speed of sound and ${l_o}$ is the length of the open pipe.
Also, the frequency of the first tone of the closed pipe is given by
${\nu _c} = \dfrac{{3v}}{{4{l_c}}}$
Here, ${\nu _c}$ is the frequency of the first tone of the closed pipe, $v$ is the speed of sound and ${l_c}$ is the length of the closed pipe.
As given in the question, the frequency of the first tone of open and closed pipe are identical, therefore, we get
${\nu _o} = {\nu _c}$
Now, putting the values of ${\nu _o}$ and ${\nu _c}$ in the above relation, we get
$\dfrac{{2\nu }}{{2{l_o}}} = \dfrac{{3\nu }}{{4{l_c}}}$
$\eqalign{
  & \Rightarrow \,\dfrac{\nu }{{{l_o}}} = \dfrac{{3\nu }}{{4{l_c}}} \cr
  & \Rightarrow \,\dfrac{1}{{{l_o}}} = \dfrac{3}{{4{l_c}}} \cr
  & \therefore \,\dfrac{{{l_c}}}{{{l_o}}} = \dfrac{3}{4} \cr} $
Hence, the lengths of open and closed pipes are in the ratio $3:4$.

So, the correct answer is “Option C”.

Note:
Here the formula for calculating the frequency of the first overtone in an open pipe is given by
$\nu = \dfrac{{mv}}{{2{l_o}}}$
Here, $m$ is the term of the harmonic wave. For the first tone, $m = 2$
Therefore, we get ${\nu _o} = \dfrac{{2v}}{{2{l_o}}}$
Now, the formula for calculating the frequency of the first overtone in a closed pipe is given by
${\nu _c} = \dfrac{{(2p - 1)v}}{{4{l_c}}}$
Where, $(2p - 1)$ represents the $nth$ harmonic.
For the closed pipe, $p = 2$, therefore, we get
 $\eqalign{
  & {\nu _c} = \dfrac{{(2(2) - 1)v}}{{4{l_c}}} \cr
  & \Rightarrow {\nu _c} = \dfrac{{\left( {4 - 1} \right)v}}{{4{l_c}}} \cr
  & \therefore \,{\nu _c} = \dfrac{{3v}}{{4{l_c}}} \cr} $
Hence, these are the terms that are used to determine the ratio of lengths of the pipe.
Recently Updated Pages