Step 1: Understanding the Question:
The question asks us to mathematically relate the fundamental acoustic frequency of an open organ pipe ($n_1$) to the fundamental frequency of a closed organ pipe ($n_2$) assuming they have identical physical lengths.
Step 2: Key Formula or Approach:
We must recall the fundamental frequency derivations for both types of organ pipes:
1. Open Pipe (open at both ends): Antinodes form at both ends. The pipe length $L$ equals half a wavelength ($\lambda/2$).
$$n_{open} = \frac{v}{\lambda} = \frac{v}{2L}$$
2. Closed Pipe (closed at one end): A node forms at the closed end, and an antinode at the open end. The pipe length $L$ equals a quarter of a wavelength ($\lambda/4$).
$$n_{closed} = \frac{v}{\lambda} = \frac{v}{4L}$$
Step 3: Detailed Explanation:
Let the fundamental frequency of the open pipe be $n_1$:
$$n_1 = \frac{v}{2L}$$
Let the fundamental frequency of the closed pipe be $n_2$:
$$n_2 = \frac{v}{4L}$$
To find the relationship, let's manipulate the equation for $n_1$ so it resembles $n_2$:
Multiply the numerator and denominator of $n_1$ by 2:
$$n_1 = \frac{2v}{4L}$$
Pull the constant 2 out to the front:
$$n_1 = 2 \times \left( \frac{v}{4L} \right)$$
Notice that the term in the parenthesis is exactly the formula for the closed pipe ($n_2$):
$$n_1 = 2 \times n_2$$
This shows that an open pipe vibrates exactly one octave higher (twice the frequency) than a closed pipe of the exact same length.
Step 4: Final Answer:
The relation is $n_1 = 2n_2$, matching option (b).