# 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). - Physics

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, = 20 cm = 0.2 m

Source frequency = nth normal mode of frequency, νn = 430 Hz

Speed of sound, v = 340 m/s

In a closed pipe, the nth 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 nth 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

Concept: Reflection of Waves - Standing Waves and Normal Modes
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Chapter 15: Waves - Exercises [Page 388]

#### APPEARS IN

NCERT Class 11 Physics
Chapter 15 Waves
Exercises | Q 17 | Page 388
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