Question

Two open organ pipes of length 20 cm are vibrating in fundamental mode simultaneously. If length of the one of the pipe reduced by 0.5 mm. Find number of beats heard per second (speed of sound 320 m/s).

Answer 1

2 Hz

Answer 2

1. Initial Frequency Calculation:

   For an open pipe, the fundamental frequency is given by

   $f = \frac{v}{2L}$

   With $L = 0.20\,\text{m}$ and $v = 320\,\text{m/s}$,

   $f_1 = \frac{320}{2 \times 0.20} = \frac{320}{0.40} = 800\,\text{Hz}$

2. New Frequency After Reduction:

   The length is reduced by $\Delta L = 0.5\,\text{mm} = 0.0005\,\text{m}$, so the new length is

   $L' = 0.20\,\text{m} - 0.0005\,\text{m} = 0.1995\,\text{m}$

   The new frequency becomes

   $f_2 = \frac{320}{2 \times 0.1995} = \frac{320}{0.399} \approx 802.0055\,\text{Hz}$

3. Beat Frequency Calculation:

   The beat frequency is the absolute difference between the two frequencies:

   $f_{\text{beat}} = |f_2 - f_1| \approx |802.0055 - 800| \approx 2\,\text{Hz}$