JEE Physics 2023 Question with Solution

Given: The first three resonance frequencies are in the ratio $1:3:5$, so the pipe behaves as a closed organ pipe. Also, $f_5 = 405 \, \text{Hz}$ and $v = 324 \, \text{m/s}^{-1}$.

Find: The length $\ell$ of the organ pipe.

For the fifth harmonic of a closed organ pipe,

$$f_5 = \frac{5v}{4\ell}$$

Substitute the given values:

$$405 = \frac{5 \times 324}{4\ell}$$

Rearranging,

$$405 \times 4\ell = 5 \times 324$$ $$1620\ell = 1620$$ $$\ell = 1 \, \text{m}$$

Therefore, the length of the organ pipe is $1 \, \text{m}$.

Given: The resonance frequencies are in the ratio $1:3:5$.

Find: Which organ pipe relation should be used.

An open organ pipe has harmonics in the ratio $1:2:3$, whereas a closed organ pipe has only odd harmonics in the ratio $1:3:5$. Therefore, this is a closed organ pipe.

Hence, the harmonic frequencies are given by

$$f_n = \frac{nv}{4\ell}, \quad n = 1,3,5,\dots$$

For the fifth harmonic,

$$f_5 = \frac{5v}{4\ell}$$

Using $f_5 = 405 \, \text{Hz}$ and $v = 324 \, \text{m/s}^{-1}$, we get

$$\ell = 1 \, \text{m}$$

So, the numerical value of the answer is 1.