Wave Motion 2 Question 44

44. The air column in a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency 440Hz. The speed of sound in air is 330m/s. End corrections may be neglected. Let p0 denote the mean pressure at any point in the pipe and Δp0 the maximum amplitude of pressure variation.

(1998,8 M)

(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?

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Answer:

Correct Answer: 44. (a) 1516m (b) ±Δp02

(c) equal to mean pressure

(d) p0+Δp0,p0Δp0

Solution:

  1. (a) Frequency of second overtone of the closed pipe

=5(v4L)=440L=5v4×440m

Substituting v= speed of sound in air =330m/s

L=5×3304×440=1516mλ=4L5=4(1516)5=34m

(b) Open end is displacement antinode. Therefore, it would be a pressure node.

 or at x=0;Δp=0

Pressure amplitude at x=x, can be written as,

Δp=±Δp0sinkx where, k=2πλ=2π3/4=8π3m1

Therefore, pressure amplitude at x=L2=15/162m or (15/32)m will be

Δp=±Δp0sin(8π3)(1532)=±Δp0sin(5π4)Δp=±Δp02

(c) Open end is a pressure node i.e. Δp=0

Hence, pmax=pmin = Mean pressure (p0)

(d) Closed end is a displacement node or pressure antinode.

 Therefore, pmax=p0+Δp0 and pmin=p0Δp0



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