Physics, asked by bunny77372, 6 months ago

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 1275 Hz source? (velocity of sound
in air = 340 m/s)

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Answered by Anonymous

Explanation:

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

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

Speed of sound, v = 340 m/s

In a closed pipe, the nth normal mode of frequency is given by the relation:

f=(2n−1)4lv n∈{1,2,3,....}

430=(2n−1)4×0.2340

2n−1=340430×4×0.2=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:

fn=nv/2l

n=2lfn/v≈0.5

The same source will not be in resonance with the same pipe open at both ends.

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A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 1275 Hz source? (velocity of soundin air = 340 m/s)​

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Hopes it help you❤️❤️

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 1275 Hz source? (velocity of sound
in air = 340 m/s)