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Question #43925

Difficulty: 8/10 | 🟢 Free

The Question

A cylindrical organ pipe of length $L$ and radius $r$ ( $r \ll L$ ) is closed at $x=0$ and open at $x=L$ . The air inside the pipe is maintained at a steady, non-uniform temperature profile $T(x) = T_0 \left(1 + \frac{x}{L}\right)^2$ , where $T_0$ is the absolute temperature at the closed end. The speed of sound in air at temperature $T_0$ is $v_0$ . The geometric end correction at the open end is given by $e_0 = 0.6r$ . Assuming the temperature variation is gradual enough that the WKB (phase integral) approximation is valid, which of the following expressions correctly represents the fundamental resonant frequency $f$ of the air column?

Options

A.
$\frac{v_0}{4(L \ln 2 + 0.3r)}$
B.
$\frac{v_0}{4(L \ln 2 + 0.6r)}$
C.
$\frac{v_0}{4(L \ln 2 + 1.2r)}$
D.
$\frac{v_0 \ln 2}{4(L + 0.6r)}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

End Correction|Adiabatic/Isothermal Processes via KTG|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the physical origin of the end correction in a non-isothermal environment.

Students typically memorize the fundamental frequency as $f = v / 4(L+e)$ and will likely integrate the velocity to find the effective length of the pipe, but blindly add $0.6r$ to this effective length (Option B).

They fail to realize that the end correction represents an acoustic mass whose phase delay depends on the local wave number at the open end.

Because the temperature is higher at the open end, the local speed of sound is doubled ( $2v_0$ ), which halves the time delay (or phase shift) contributed by the geometric end correction, effectively making it behave as $0.3r$ relative to the base velocity $v_0$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	End Correction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free