Question

The pressure wave in a closed organ pipe of length $40\text{cm}$ is given by $p=5\sin \left(\frac{2\pi x}{32}\right)\cos \left(50\pi t\right)$ where $x,p$ and $t$ are in $\text{cm},{\text{N/m}}^2$ and seconds respectively. The pressure nodes are located at which of the following distances along the pipe (from the open end) ?

• A. $8\text{cm},24\text{cm}$
• B. $16\text{cm},32\text{cm}$
• C. $16\text{cm},24\text{cm}$
• D. $18\text{cm},32\text{cm}$

Answer 1

The correct option is B $16\text{cm},32\text{cm}$

Given,
The pressure wave is
$p=5\sin \left(\frac{2\pi x}{32}\right)\cos \left(50\pi t\right)$

General equation for pressure wave in a closed organ pipe is

$p=-2p_o\cos \left(\frac{2\pi }{\lambda }x\right)\sin \left(\frac{2\pi }{T}t\right)$

Comparing both equations, we get

$k=\frac{2\pi }{\lambda }=\frac{2\pi }{32}$

$\Rightarrow \lambda =32\text{cm}$

Positions of zero pressure variation (or pressure nodes) are given by
$x=\frac{n\lambda }{2}\text{where}n=0,1,2,\mathrm{3....}$
$\therefore x=0,\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2}......$
$\because \lambda =32\text{cm}$
we can conclude that,
For, $n=0,x=0\text{cm}$
$n=1,x=16\text{cm}$
$n=2,x=32\text{cm}$
$n=3,x=48\text{cm}$

So, among the given options, only option (b) satisfies the position of pressure node.