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 $440 Hz$. The speed of sound in air is $330 m / s$. End corrections may be neglected. Let $p _0$ denote the mean pressure at any point in the pipe and $\Delta p _0$ 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) $\frac{15}{16} m$ (b) $\pm \frac{\Delta p _0}{\sqrt{2}}$
(c) equal to mean pressure
(d) $p _0+\Delta p _0, p _0-\Delta p _0$
Solution:
- (a) Frequency of second overtone of the closed pipe
$$ =5\left(\frac{v}{4 L}\right)=440 \Rightarrow L=\frac{5 v}{4 \times 440} m $$
Substituting $v=$ speed of sound in air $=330 m / s$
$$ \begin{aligned} & L=\frac{5 \times 330}{4 \times 440}=\frac{15}{16} m \\ & \lambda=\frac{4 L}{5}=\frac{4\left(\frac{15}{16}\right)}{5}=\frac{3}{4} m \end{aligned} $$
(b) Open end is displacement antinode. Therefore, it would be a pressure node.
$$ \text { or at } \quad x=0 ; \Delta p=0 $$
Pressure amplitude at $x=x$, can be written as,
$$ \begin{aligned} \Delta p & = \pm \Delta p _0 \sin k x \\ \text { where, } \quad k & =\frac{2 \pi}{\lambda}=\frac{2 \pi}{3 / 4}=\frac{8 \pi}{3} m^{-1} \end{aligned} $$
Therefore, pressure amplitude at $x=\frac{L}{2}=\frac{15 / 16}{2} m$ or $(15 / 32) m$ will be
$$ \begin{aligned} \Delta p & = \pm \Delta p _0 \sin \left(\frac{8 \pi}{3}\right)\left(\frac{15}{32}\right) \\ & = \pm \Delta p _0 \sin \left(\frac{5 \pi}{4}\right) \\ \Delta p & = \pm \frac{\Delta p _0}{\sqrt{2}} \end{aligned} $$
(c) Open end is a pressure node i.e. $\Delta p=0$
Hence, $p _{\max }=p _{\text {min }}=$ Mean pressure $\left(p _0\right)$
(d) Closed end is a displacement node or pressure antinode.
$$ \begin{array}{ll} \text { Therefore, } & p _{\max }=p _0+\Delta p _0 \\ \text { and } & p _{\min }=p _0-\Delta p _0 \end{array} $$
