If n1 is the resonance frequency of a pipe open at both ends and n2 the resonance frequency of pipe open at one ends only, and both are vibrating in the fundamental mode and the pipes are of the same length, then (a) n1 = 2n2 (b) n1 = n2 (c) 2n1 = n2 (d) 3n2 = 4n2

To find the relationship between the resonance frequencies of a pipe open at both ends and a pipe open at one end, let’s break down the concepts of resonance frequency and the modes of vibration for each type of pipe.

Understanding Resonance Frequencies

Resonance frequency is the frequency at which a system naturally oscillates. For pipes, the mode of vibration depends on whether they are open at both ends or closed at one end.

Pipes Open at Both Ends

A pipe that is open at both ends supports standing waves with nodes (points of no movement) at the open ends. In the fundamental mode (first harmonic), the wavelength (λ) is twice the length (L) of the pipe:

• λ1 = 2L

The relationship between the speed of sound (v), frequency (n1), and wavelength (λ) gives us:

• v = n1 * λ1

Substituting for λ1:

• v = n1 * 2L
• n1 = v / (2L)

Pipes Open at One End

For a pipe that is closed at one end (and open at the other), it supports standing waves with a node at the closed end and an antinode at the open end. In the fundamental mode, the wavelength (λ) is four times the length of the pipe:

• λ2 = 4L

Using the same relationship for frequency:

• v = n2 * λ2

Again, substituting for λ2 gives us:

• v = n2 * 4L
• n2 = v / (4L)

Establishing the Relationship

Now, we have both frequencies expressed in terms of the speed of sound and the length of the pipes:

• n1 = v / (2L)
• n2 = v / (4L)

To compare n1 and n2, we can express n2 in terms of n1:

• From n2: n2 = v / (4L) = (1/2) * (v / (2L)) = (1/2) * n1

Final Relationship

This leads us to the conclusion that:

• n1 = 2n2

So, the correct answer is (a) n1 = 2n2. This means that the fundamental frequency of a pipe open at both ends is twice that of a pipe open at one end, given they are of the same length.