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Question:
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).

Answer:

First (Fundamental); No

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

Source frequency = n^th normal mode of frequency, V_n = 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 integer = 0, 1, 2, 3...

430 = (2n-1) 340 / (4x0.2)

2n -1 = (430 x 4 x 0.2) / 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

= 2 x 0.2 x 430 / 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.

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Question:
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).

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Question: 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).

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Question:
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).