
A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. Determine the ratio of their lengths.
A. 1 : 1
B. 2 : 1
C. 1 : 4
D. 1 : 2
Answer
142.8k+ views
Hint: In this question, we need to find the ratio of closed and open organ pipes if they are tuned to the same frequency. So, we need to use the following formula. After, equating the equations for closed organ pipe and open organ pipe, we will get the desired result.
Formula used:
The formula for fundamental frequency for closed organ pipe is given by
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Similarly, the formula for fundamental frequency for open organ pipe is given by
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Where, \[{f_c}\] is the fundamental frequency for closed organ pipe, \[{f_o}\] is the fundamental frequency for open organ pipe, \[v\] is the speed of wave, \[{L_c}\] is the length of closed organ pipe and \[{L_o}\] is the length of open organ pipe.
Complete step by step solution:
We know that, the basic frequency for closed organ pipe is,
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Also, the basic frequency for closed organ pipe is,
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Here, speed is constant.
So, according to the given condition, both the pipes are tuned to the same frequency.Thus, we get
\[{f_c} = {f_o}\]
So, \[\dfrac{v}{{4{L_c}}} = \dfrac{v}{{2{L_o}}}\]
By simplifying, we get
\[\dfrac{1}{{4{L_c}}} = \dfrac{1}{{2{L_o}}}\]
\[\Rightarrow 4{L_c} = 2{L_o}\]
\[\Rightarrow 2{L_c} = {L_o}\]
By simplifying, further, we get
\[\dfrac{{{L_c}}}{{{L_o}}} = \dfrac{1}{2}\]
That means \[{L_c}:{L_o} = 1:2\]
Hence, the ratio of closed and open organ pipes, if they are tuned to the same frequency, is 1:2.
Therefore, the correct option is (D).
Note: Many students make mistakes in writing the formula for the fundamental frequency of a pipe. Consequently, the end result may get wrong. Here, the simplification part is also important for getting the final answer.
Formula used:
The formula for fundamental frequency for closed organ pipe is given by
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Similarly, the formula for fundamental frequency for open organ pipe is given by
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Where, \[{f_c}\] is the fundamental frequency for closed organ pipe, \[{f_o}\] is the fundamental frequency for open organ pipe, \[v\] is the speed of wave, \[{L_c}\] is the length of closed organ pipe and \[{L_o}\] is the length of open organ pipe.
Complete step by step solution:
We know that, the basic frequency for closed organ pipe is,
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Also, the basic frequency for closed organ pipe is,
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Here, speed is constant.
So, according to the given condition, both the pipes are tuned to the same frequency.Thus, we get
\[{f_c} = {f_o}\]
So, \[\dfrac{v}{{4{L_c}}} = \dfrac{v}{{2{L_o}}}\]
By simplifying, we get
\[\dfrac{1}{{4{L_c}}} = \dfrac{1}{{2{L_o}}}\]
\[\Rightarrow 4{L_c} = 2{L_o}\]
\[\Rightarrow 2{L_c} = {L_o}\]
By simplifying, further, we get
\[\dfrac{{{L_c}}}{{{L_o}}} = \dfrac{1}{2}\]
That means \[{L_c}:{L_o} = 1:2\]
Hence, the ratio of closed and open organ pipes, if they are tuned to the same frequency, is 1:2.
Therefore, the correct option is (D).
Note: Many students make mistakes in writing the formula for the fundamental frequency of a pipe. Consequently, the end result may get wrong. Here, the simplification part is also important for getting the final answer.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Physics Average Value and RMS Value JEE Main 2025

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
