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

Question

Two open pipes of length L are vibrated simultaneously. If length of one of the pipe is reduced by y, then the number of beats heard per second will be. (if the velocity of sound is V and y < < L)

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Beat frequency due to change in frequency of open pipes
	**Related Topics**: Frequency of open pipes, Beat frequency
	**Prerequisites**: Understanding of open pipe harmonics, Beat frequency definition
## SOLUTION PATHWAY
**Given Information**:
	* Two open pipes of length L.
	* Length of one pipe reduced by y.
	* Velocity of sound = V
	* y << L
**Method & Steps**:
	1. Find the fundamental frequency of the original pipes:
		Reasoning: The fundamental frequency of an open pipe is given by $f = \frac{v}{2L}$, where v is the speed of sound and L is the length of the pipe.
		Formula/Rule: $f = \frac{v}{2L}$
		Working: $f_1 = \frac{V}{2L}$
	2. Find the fundamental frequency of the pipe with reduced length:
		Reasoning: Length is now L - y. Let this frequency be f2.
		Working: $f_2 = \frac{V}{2(L-y)}$
	3.  Calculate the beat frequency:
		Reasoning: Beat frequency is the absolute difference between the two frequencies.  $f_{beat} = |f_2 - f_1|$
		Working: $f_{beat} = |\frac{V}{2(L-y)} - \frac{V}{2L}| = |\frac{VL - V(L-y)}{2L(L-y)}| = |\frac{VL - VL + Vy}{2L(L-y)}| = \frac{Vy}{2L(L-y)}$
	4.  Apply the approximation y << L:
		Reasoning: Since y << L, we can approximate (L-y) ≈ L
		Working: $f_{beat} \approx \frac{Vy}{2L^2}$  
	5.  Re-evaluate the equation using approximation
		Reasoning: In the approximated equation, The approximation should be made to (L-y) ≈ L and not to L^2
		Working: $f_{beat} \approx \frac{Vy}{2L(L)} = \frac{Vy}{2L^2}$ is incorrect, the correct approximation is $f_{beat} = \frac{Vy}{2L(L-y)} \approx \frac{Vy}{2L(L)} = \frac{Vy}{2L}$ 
		Key Point: The approximation is (L-y) ≈ L, not (L-y) ≈ 0
	3. Final Result:
		Result: $f_{beat} = \frac{Vy}{2L}$
		Verification: Dimensional analysis shows units are correct: (m/s * m) / m = 1/s which is frequency
 ## OPTION ANALYSIS
[A] :x:  This is incorrect as it has a L^2 term in the denominator, which is not correct.
[B] :x: This is incorrect as it misses the factor of 2 in the denominator.
[C] ✓: This is the correct beat frequency using the approximation.
[D] :x: This is incorrect as it has an inverse relationship of length and speed with frequency.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the beat frequency is the difference of frequencies.
			[How to avoid]: Remember the definition of beat frequency.
		2.[Error 2]: Incorrectly applying the approximation.
			[How to avoid]: Make sure the approximation is applied to the length and not to length squared or other terms.
	**Quick Tips**:
		1.[Helpful shortcut]: If y << L, then (L-y) ≈ L can be used to simplify calculations.
		2.[Memory aid]: Beat frequency is the absolute difference of the two frequencies.

SUBHANI · Answer

$\frac{{Vy}}{{2{L^2}}}$

SUBHANI · Answer

$\frac{{Vy}}{{{L^2}}}$

SUBHANI · Answer

The correct Option is "$\frac{{Vy}}{{2L}}$"

SUBHANI · Answer

$\frac{{2{L^2}}}{{Vy}}$

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