Question

Industrial espionage on a distance planet (temperature and atmosphere is unknown) provides the information that successive resonances in a pipe are 210 and 294 Hz. The pipe is 5 m long, but it’s not known whether or not the pipe is open at both ends or only at one end. What is the pipe’s fundamental resonance frequency? The answer is 42Hz

Answer 1

L= length of pipe = 5m

For open tube, frequencies are-

v/2L, 2(v/2L), 3(v/2L)……..n(v/2L), (n+1) (v/2L)

For closed tube, frequencies are-

v/4L, 3(v/4L), 5(v/4L)……..(2n-1)(v/4L), (2n+1) (v/4L)

given successive resonances are – 210 Hz and 294Hz

Let’s check if pipe is open or closed.

1. For open- n(v/2L) / (n+1) (v/2L) = 210/ 294

n/ (n+1) = 210/294

n= 2.5

2. for closed-   (2n-1)(v/4L) / (2n+1) (v/4L) = 210/ 294

                (2n-1) / (2n+1) = 210/ 294

                          n=3

as we get whole number n=3 so tube is closed.

(2n-1)(v/4L) - (2n+1) (v/4L) = 210 – 294

                 2 (v/4L) = 84

                          v/4L = 42 Hz

fundamental frequency of closed pipe = v/4L = 42 Hz