Question

An air column in a tube of length 50 cm, closed at one end is vibrating in its fifth harmonic. The phase difference between a particle at the open end and a particle at 42 cm from the open end is

• A. \( 90^\circ \)
• B. \( 180^\circ \)
• C. \( 0^\circ \)
• D. \( 270^\circ \)

Hint:

In a standing wave, all points between any two adjacent nodes are in phase. The phase flips by \(180^\circ\) (or \(\pi\) radians) every time you cross a node. By locating the nodes, you can quickly determine the relative phase of any two points.

Answer 1

The Correct Option is C

For a tube closed at one end, the possible wavelengths are given by \( L = (2n-1)\frac{\lambda}{4} \), where L is the length of the tube and n is the harmonic number (n=1, 3, 5,... for closed pipes). The question says "fifth harmonic", which corresponds to the harmonic number n=3 in the series of possible harmonics for a closed pipe (1st, 3rd, 5th...). Let's assume "fifth harmonic" means \(n=5\). This would be the 9th overtone which is unusual. Let's assume it means the 5th possible resonant mode.
Let's take "fifth harmonic" to mean the mode with 5 quarter-wavelengths. For a closed pipe, the nth harmonic has a frequency of \( (2n-1)f_1 \). The 5th harmonic would be for n=3. Let's assume the question means n=5. Then the mode is the 9th overtone. This is likely a misstatement, and it means the 2nd overtone, which is the 5th harmonic for an open pipe. Let's assume it's the 5th natural frequency for a closed pipe, which is \( n=3 \) in the formula \(f_n = (2n-1)f_1\). Wait, the harmonics are named by the multiple of the fundamental. So 5th harmonic is \(5 \times f_1\). A closed pipe only has odd harmonics. So this is the 5th harmonic.
The length L contains \( (2n-1) \) quarter wavelengths. So for the 5th harmonic, \(n=3\) is not right. For the nth harmonic \(f_n = n f_1\). For a closed pipe, only odd n are possible. So the 5th harmonic is when n=5.
So for the 5th harmonic, \( L = 5 \frac{\lambda}{4} \).
Given \( L = 50 \) cm = 0.5 m.
\( 0.5 = 5 \frac{\lambda}{4} \implies \lambda = \frac{0.5 \times 4}{5} = \frac{2}{5} = 0.4 \) m = 40 cm.
The phase difference \( \Delta\phi \) between two points separated by a distance \( \Delta x \) is given by \( \Delta\phi = \frac{2\pi}{\lambda} \Delta x \).
One point is at the open end (\(x=0\)). The other point is at \(x = 42\) cm from the open end.
The distance between them is \( \Delta x = 42 \) cm.
Let's calculate the phase difference.
\( \Delta\phi = \frac{2\pi}{40 \text{ cm}} \times 42 \text{ cm} = \frac{84\pi}{40} = \frac{21\pi}{10} = 2.1\pi \).
This isn't a simple angle. Let's re-read the question. A standing wave is formed.
In a standing wave, all particles between two consecutive nodes vibrate in phase. And particles in adjacent segments (separated by a node) vibrate in opposite phase (\(180^\circ\) difference).
The open end is an antinode. The closed end is a node. The positions of the nodes from the open end are at \( x = \lambda/4, 3\lambda/4, 5\lambda/4, ... \).
With \( \lambda = 40 \) cm, the nodes are at:
\(x_1 = 40/4 = 10\) cm.
\(x_2 = 3(40)/4 = 30\) cm.
\(x_3 = 5(40)/4 = 50\) cm (at the closed end).
The antinodes are at \(x=0, \lambda/2, \lambda, ...\), which are 0 cm, 20 cm, 40 cm.
The points are at the open end (\(x=0\)) and at \(x=42\) cm.
The segment between node at 30cm and node at 50cm contains the point \(x=42\)cm. The antinode for this segment is at 40cm.
The point at the open end (x=0) is in the first segment (between x=0 and node at x=10). The point at x=42cm is in the third segment (between node at x=30 and node at x=50).
The phase relationship is: Segment 1 (0-10cm): phase \(\phi_0\) Segment 2 (10-30cm): phase \(\phi_0 + \pi\) Segment 3 (30-50cm): phase \(\phi_0 + 2\pi\), which is the same as \(\phi_0\).
Since the point at x=0 (open end) and the point at x=42 cm are in segments with the same phase, the phase difference between them is 0 or \(360^\circ\).
Therefore, the phase difference is \(0^\circ\).