a) Define an organ pipe. [1] b) What is the ratio of length of two pipes, an open organ pipe A and a closed organ pipe B both vibrating in third overtones are in resonance? Justify. [2] c) In the…

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a) Define an organ pipe.

An organ pipe is a hollow cylindrical tube, open at one or both ends, used to produce musical notes by generating standing waves of sound. Organ pipes are classified into two main types based on their structure and the way they produce sound:

1. Open organ pipe (or open-ended pipe): Both ends of the pipe are open. The fundamental frequency (first harmonic) is produced when the length of the pipe is half the wavelength of the sound wave.
2. Closed organ pipe (or closed-ended pipe): One end of the pipe is closed, and the other end is open. The fundamental frequency is produced when the length of the pipe is one-fourth the wavelength of the sound wave.

The vibrations of air inside the pipe create standing waves, which determine the pitch of the sound produced.

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b) What is the ratio of length of two pipes, an open organ pipe A and a closed organ pipe B both vibrating in third overtones are in resonance? Justify.

Step 1: Understand the harmonics in open and closed pipes

• Open organ pipe (A):

  • Both ends are open, so it supports all harmonics (n = 1, 2, 3, ...).
  • The length $L_A$ of the pipe for the $n$-th harmonic is given by: $$L_A=\frac{n{\lambda }_A}{2}$$ where ${\lambda }_A$ is the wavelength of the sound produced.

• Closed organ pipe (B):

  • One end is closed, so it supports only odd harmonics (n = 1, 3, 5, ...).
  • The length $L_B$ of the pipe for the $n$-th harmonic is given by: $$L_B=\frac{(2n-1){\lambda }_B}{4}$$ where ${\lambda }_B$ is the wavelength of the sound produced.

Step 2: Define the third overtone

• The third overtone corresponds to the fourth harmonic (since the fundamental is the first harmonic, the first overtone is the second harmonic, and so on).

  • For the open pipe (A), the third overtone is the 4th harmonic ($n=4$).
  • For the closed pipe (B), the third overtone is the 5th harmonic ($n=5$ for odd harmonics, but since the third overtone is the 4th harmonic in terms of counting, we need to clarify: the third overtone in a closed pipe is the 5th harmonic because the closed pipe only supports odd harmonics. However, the question states "third overtone," which typically means the 4th harmonic in general terms. But in closed pipes, the third overtone is the 5th harmonic (since the first overtone is the 3rd harmonic, the second overtone is the 5th harmonic, and the third overtone is the 7th harmonic). This is a common point of confusion.

  Clarification:

  • In open pipes, the harmonics are $n=1,2,3,4,\dots$. The third overtone is the 4th harmonic ($n=4$).
  • In closed pipes, the harmonics are $n=1,3,5,7,\dots$. The third overtone is the 7th harmonic ($n=7$, since the first overtone is the 3rd harmonic, the second overtone is the 5th harmonic, and the third overtone is the 7th harmonic).

  However, the question states "third overtone," which is ambiguous. For consistency with standard definitions:

  • Open pipe (A): Third overtone = 4th harmonic ($n=4$).
  • Closed pipe (B): Third overtone = 7th harmonic ($n=7$).

  But let's assume the question means the third harmonic (i.e., the third resonance mode, not the third overtone). This is a common interpretation in such problems. For an open pipe, the third harmonic is $n=3$, and for a closed pipe, the third harmonic is $n=3$ (but closed pipes only have odd harmonics, so the third harmonic is $n=3$).

  Revised Interpretation: The question likely means the third harmonic (not the third overtone). This is more standard in such problems. So:

  • For the open pipe (A), the third harmonic is $n=3$.
  • For the closed pipe (B), the third harmonic is $n=3$ (since closed pipes only have odd harmonics: 1st, 3rd, 5th, etc.).

  However, the question explicitly says "third overtone," so we proceed with that.

  Final Interpretation:

  • Open pipe (A): Third overtone = 4th harmonic ($n=4$).
  • Closed pipe (B): Third overtone = 7th harmonic ($n=7$).

Step 3: Write expressions for the lengths

For the open pipe (A) vibrating in the 4th harmonic: $$L_A=\frac{4{\lambda }_A}{2}=2{\lambda }_A$$

For the closed pipe (B) vibrating in the 7th harmonic: $$L_B=\frac{(2\times 7-1){\lambda }_B}{4}=\frac{13{\lambda }_B}{4}$$

Step 4: Condition for resonance

For the two pipes to be in resonance, the frequencies of the sound produced must be equal: $$f_A=f_B$$ Since $f=\frac{v}{\lambda }$, where $v$ is the speed of sound (same for both pipes in air): $$\frac{v}{{\lambda }_A}=\frac{v}{{\lambda }_B}⟹{\lambda }_A={\lambda }_B$$

Step 5: Find the ratio $\frac{L_A}{L_B}$

Substitute ${\lambda }_A={\lambda }_B$ into the length equations: $$L_A=2{\lambda }_A$$ $$L_B=\frac{13{\lambda }_A}{4}$$

Now, find the ratio: $$\frac{L_A}{L_B}=\frac{2{\lambda }_A}{\frac{13{\lambda }_A}{4}}=\frac{2\times 4}{13}=\frac{8}{13}$$

However, this seems inconsistent with standard interpretations. Let's re-examine the definition of "third overtone."

Correct Interpretation of "Third Overtone"

The third overtone is the fourth harmonic in general terms. For a closed pipe, the harmonics are odd, so:

• First overtone: 3rd harmonic ($n=3$)
• Second overtone: 5th harmonic ($n=5$)
• Third overtone: 7th harmonic ($n=7$)

For an open pipe:

• First overtone: 2nd harmonic ($n=2$)
• Second overtone: 3rd harmonic ($n=3$)
• Third overtone: 4th harmonic ($n=4$)

Thus, the correct interpretation is:

• Open pipe (A): Third overtone = 4th harmonic ($n=4$)
• Closed pipe (B): Third overtone = 7th harmonic ($n=7$)

Now, the lengths are: $$L_A=\frac{4{\lambda }_A}{2}=2{\lambda }_A$$ $$L_B=\frac{(2\times 7-1){\lambda }_B}{4}=\frac{13{\lambda }_B}{4}$$

For resonance, ${\lambda }_A={\lambda }_B$, so: $$\frac{L_A}{L_B}=\frac{2{\lambda }_A}{\frac{13{\lambda }_A}{4}}=\frac{8}{13}$$

But this seems counterintuitive because the closed pipe's length should be shorter for higher harmonics. Let's verify the harmonic numbering again.

Alternative Approach: Using Harmonic Numbers

For an open pipe, the frequency of the $n$-th harmonic is: $$f_n=\frac{nv}{2L_A}$$

For a closed pipe, the frequency of the $m$-th harmonic (where $m$ is odd) is: $$f_m=\frac{mv}{4L_B}$$

For resonance, $f_n=f_m$: $$\frac{nv}{2L_A}=\frac{mv}{4L_B}⟹\frac{n}{L_A}=\frac{m}{2L_B}$$

For the third overtone:

• Open pipe: $n=4$ (since first overtone is $n=2$, second is $n=3$, third is $n=4$)
• Closed pipe: $m=7$ (since first overtone is $m=3$, second is $m=5$, third is $m=7$)

Substitute $n=4$ and $m=7$: $$\frac{4}{L_A}=\frac{7}{2L_B}⟹\frac{L_A}{L_B}=\frac{8}{7}$$

This makes more sense. The ratio of the lengths is $\frac{8}{7}$.

Final Calculation:

$$\frac{L_A}{L_B}=\frac{8}{7}$$

Thus, the ratio of the length of the open pipe (A) to the closed pipe (B) is 8:7.

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c) In the given figure below, figure (i) represents a situation that both source of sound (S) and Listener (L) are at rest position 'S' and 'L' respectively. Figure (ii) represents relative motion between moving source and stationary listener.

i) Does the pitch of sound changes to the listener in the second case? Justify.

Answer: Yes, the pitch of the sound changes to the listener in the second case (when the source is moving and the listener is stationary). This phenomenon is known as the Doppler effect.

Justification:

The Doppler effect describes the change in frequency (pitch) of a wave for an observer moving relative to its source. When the source of sound moves toward a stationary listener:

1. The wavelength of the sound in front of the source is compressed (shorter wavelength, ${\lambda }^{\prime }<\lambda$).
2. The wavelength behind the source is stretched (longer wavelength, ${\lambda }^{\prime \prime }>\lambda$).
3. The frequency perceived by the listener increases because the waves reach the listener more frequently (higher pitch).

Mathematically, the observed frequency $f^{\prime }$ when the source is moving toward the listener is given by: $$f^{\prime }=\frac{v}{v-v_s}f$$ where:

• $v$ is the speed of sound in the medium,
• $v_s$ is the speed of the source,
• $f$ is the original frequency of the sound.

Since $v_s>0$ (source moving toward the listener), $f^{\prime }>f$, meaning the pitch increases.

In the given figure:

• In Figure (i), both the source and listener are stationary, so no Doppler shift occurs, and the pitch remains unchanged.
• In Figure (ii), the source is moving toward the stationary listener, causing a higher pitch due to the Doppler effect.