Question

For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:5 respectively. If the frequency of fifth harmonic is 405 Hz and the speed of sound in air is 324 ms^-1 the length of the organ pipe is ___ m.

Answer 1

Correct Answer: 1

Solution:
For the 5th harmonic in a closed organ pipe, the relationship between the frequency \( f \), speed of sound \( v \), and the length \( \ell \) of the pipe is given by: \[ f_5 = \frac{5v}{4\ell} \] Given:
- \( f_5 = 405 \, \text{Hz} \)
- \( v = 324 \, \text{m/s}^{-1} \)

Substitute the given values into the equation: \[ 405 = \frac{5 \times 324}{4\ell} \] Solving for \( \ell \): \[ 405 \times 4\ell = 5 \times 324 \] \[ 1620\ell = 1620 \] \[ \ell = 1 \, \text{m} \] Thus, the length of the organ pipe is \( \boxed{1} \, \text{m} \).