An organ pipe has one end closed and at the other end is a vibrating diaphragm, which is a displacement antinode. When the frequency of the diaphragm is $2,000 \mathrm{~Hz}$ a stationary or standing wave pattern is set up in the tube. The distance between adjacent nodes is $8.0 \mathrm{~cm}$. As the frequency is slowly reduced the stationary wave pattern disappears, but another stationary wave pattern reappears at frequency 1,600 Hz. Calculate:
(i) the speed of sound in air
(ii) the distance between adjacent nodes at $1,600 \mathrm{~Hz}$
(iii) the length of the tube
(iv) the next frequency below $1,600 \mathrm{~Hz}$ at which a stationary pattern occurs. [6]
Show answer · 显示答案
- Because consecutive resonances are always $\dfrac{v}{2L}$ apart, the data fix both $L = 0.40\,\text{m}$ (part iii) and the next pattern below $1600\,\text{Hz}$, at $1200\,\text{Hz}$.
Worked solutions are in the book.


