if v1 is the resonance frequency of a pipe open at both ends and v2 the resonance frequency of a pipe open at, one end only and both are vibrating in the fundamental mode, and the pipes are of same length, then

When we talk about resonance frequencies in pipes, we need to consider how the boundaries of the pipes affect the standing waves that form within them. In your question, you have two types of pipes: one that is open at both ends (let's call it Pipe A) and another that is open at one end (Pipe B). Both pipes are of the same length and vibrating in their fundamental mode. Let's break down the resonance frequencies of each pipe.

Understanding Resonance Frequencies

Resonance frequency is the frequency at which a system naturally oscillates. For pipes, this frequency depends on how many nodes and antinodes can form based on the pipe's boundary conditions.

Pipe Open at Both Ends

For Pipe A, which is open at both ends, the fundamental mode (first harmonic) has a node at the center and antinodes at both ends. The length of the pipe is equal to half the wavelength of the sound wave produced. Mathematically, this can be expressed as:

• Length of Pipe A (L) = λ/2
• Resonance frequency (v1) = f × λ

From the relationship between frequency, wavelength, and speed of sound (v = f × λ), we can derive that:

• λ = 2L
• v1 = f × 2L

Pipe Open at One End

Now, for Pipe B, which is open at one end and closed at the other, the fundamental mode has a node at the closed end and an antinode at the open end. In this case, the length of the pipe is equal to a quarter of the wavelength:

• Length of Pipe B (L) = λ/4
• Resonance frequency (v2) = f × λ

Using the same relationship, we find:

• λ = 4L
• v2 = f × 4L

Comparing the Frequencies

Now that we have the expressions for both resonance frequencies, we can compare them. From our earlier derivations:

• v1 = f × 2L
• v2 = f × 4L

To find the relationship between v1 and v2, we can express them in terms of the same fundamental frequency (f):

• v1 = 2fL
• v2 = 4fL

From this, we can see that:

• v2 = 2 × v1

Final Thoughts

This means that the resonance frequency of the pipe open at one end (v2) is twice that of the pipe open at both ends (v1). This relationship highlights how the boundary conditions of a pipe significantly influence its resonance frequencies. In practical terms, this is why you might notice that a flute (which is open at both ends) produces different pitches compared to a clarinet (which is closed at one end) even when they are of similar lengths.