A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s^{–1}).

#### Solution 1

First (Fundamental); No

Length of the pipe, *l *= 20 cm = 0.2 m

Source frequency = *n*^{th} normal mode of frequency, ν_{n}_{ }= 430 Hz

Speed of sound, *v* = 340 m/s

In a closed pipe, the *n*^{th} normal mode of frequency is given by the relation:

`v_n = (2n - 1) v/(4l)` n is an interger = 0,1.2.3

`430 =(2n - 1) 340/(4xx0.2)`

`2n -1 = (430xx4xx0.2)/340 = 1.01`

2n = 2.01

n ~ 1

Hence, the first mode of vibration frequency is resonantly excited by the given source. In a pipe open at both ends, the *n*^{th} mode of vibration frequency is given by the relation:

`v_n = "nv"/(2l)`

`n = (2lv_n)/v`

`= (2xx0.2xx430)/340 = 0.5`

Since the number of the mode of vibration (*n*) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

#### Solution 2

Here length of pipe, 1 = 20 cm = 0.20 m, frequency v = 430 Hz and speed of sound in air υ = 340 ms^{-1}

For closed end pipe, `v = ((2n - 1)v)/(4l)` where n = 1, 2, 3.....

`:. (2n -1) = (4vl)/v = (4xx430xx0.20)/340 = 1.02`

`=> 2n = 1.02 + 1= 2.02 => n = 0.20/2 = 1.01`

Hence, resonance can occur only for first (or fundamental) mode of vibration.

As for an open pipe ` v = (nv)/(2l)` where n = 1, 2,3 ....

`:. n = (2lv)/v = (2xx430xx0.20)/(340) = 0.51`

As n < 1, hence in this case resonance position cannot be obtained