The Question is from neet-physics waves , and it was created by the author VENKATESWARARAO

Question

The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length $2\,\,m$. length of the open pipe is

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Resonance in open and closed pipes
	**Related Topics**: Standing waves, harmonics, fundamental frequency, overtones
	**Prerequisites**: Understanding of wave properties, resonance, and the behavior of sound waves in open and closed pipes.
## SOLUTION PATHWAY
**Given Information**:
	- Length of the closed pipe, $L_c = 2$ m
	- Second overtone of open pipe resonates with the first overtone of the closed pipe.

**Method & Steps**:
	1. **Determine the frequency of the first overtone of the closed pipe:**
		Reasoning: In a closed pipe, the first overtone corresponds to the third harmonic. The frequency of the nth harmonic of a closed pipe is given by $f_n = n \frac{v}{4L_c}$, where $n$ is an odd integer (1, 3, 5,...), $v$ is the speed of sound, and $L_c$ is the length of the closed pipe.
		Formula/Rule: $f_3 = 3 \frac{v}{4L_c}$
	2. **Calculate the frequency of the first overtone of the closed pipe:**
		Working: $f_3 = 3 \frac{v}{4(2)} =  \frac{3v}{8}$
		Key Point: This is the resonant frequency.
	3. **Determine the frequency of the second overtone of the open pipe:**
		Reasoning: In an open pipe, the second overtone corresponds to the third harmonic. The frequency of the nth harmonic of an open pipe is given by $f_n = n \frac{v}{2L_o}$, where $n$ is an integer (1, 2, 3,...), $v$ is the speed of sound, and $L_o$ is the length of the open pipe.
		Formula/Rule: $f_3 = 3 \frac{v}{2L_o}$
	4. **Equate the frequencies and solve for the length of the open pipe:**
		Working: $\frac{3v}{8} = 3 \frac{v}{2L_o}$  => $\frac{1}{8} = \frac{1}{2L_o}$ => $2L_o = 8$ => $L_o = 4$ m
		Result: The length of the open pipe is 4 m.
		Verification:  The second overtone of the 4m open pipe has the same frequency as the first overtone of the 2m closed pipe.
## OPTION ANALYSIS
[A] ✓: The length of the open pipe is calculated to be 4 m, matching the correct answer.
[B] :x: This would not satisfy the resonance condition.
[C] :x: This would not satisfy the resonance condition.
[D] :x: This would not satisfy the resonance condition.
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing harmonics and overtones: Remember that the first overtone is the second harmonic, the second overtone is the third harmonic, etc.
		2. Using the wrong formula for open and closed pipes: Open pipes have harmonics at all integer multiples of the fundamental frequency, while closed pipes only have odd integer multiples.
	**Quick Tips**:
		1. Draw a diagram of the waves in the pipes to visualize the harmonics.
		2. Use the general formula $f = n\frac{v}{2L}$ for open pipes and $f = n\frac{v}{4L}$ for closed pipes, keeping in mind the allowed values of n.

VENKATESWARARAO · Answer

The correct Option is "$4\,\,m$"

VENKATESWARARAO · Answer

$2\,\,m$

VENKATESWARARAO · Answer

$8\,\,m$

VENKATESWARARAO · Answer

$1\,\,m$

NEET Physics Waves MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company