The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 300 m s -1 . End corrections may be neglected. Let P 0 denote the mean pressure at any point in the pipe, and ∆P 0 the maximum amplitude of pressure variation. (a) Find the length L of the air column. (b) What is the amplitude of pressure variation at the middle of the column? (c) What are the maximum and minimum pressures at the open end of the pipe ? (d) What are the maximum and minimum pressures at the closed end of the pipe ?

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

(a) For second overtone as shown,

5ℷ/4 = ℓ ∴ ℷ = 4ℓ/5

Also, v = vℷ

=> 330 = 440 x 4ℓ/5 => ℓ = 15/16 m.

(b) KEY CONCEPT : At any position x, the pressure is given by

∆P = ∆P₀ cos kx cos ω t

Here amplitude A = ∆P₀ cos kx = ∆P₀ cos 2π/ℷ x

For x = 15/2 x 16 = 15/32 m (mid point)

Amplitude = ∆P₀ cos [2π/(330 / 440) x 15/32] = ∆P₀/√2

(c) At open end of pipe, pressure is always same i.e. equal to mean pressure

∵ ∆P = 0, P_max = P_min = P₀

(d) At the closed end :

Maximum Pressure = P₀ + ∆P₀

Minimum Pressure = P₀ - ∆P₀

Thanks
Aditi Chauhan
askIITians Faculty

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Wave Motion

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship