Question

The equation of a standing wave in an open organ pipe is given by: $p=16\sin \left(\frac{\pi }{3}x\right)\cos \left(64t\right)$, where symbols have their usual meaning. Here $x$ is in $\text{cm}$, $t$ is in $\text{seconds}$ and the amplitude is given in $\text{cm}$ .Then, the first pressure node is obtained at a distance $x$ from one end. What is the value of $x$?

• A. $2\text{cm}$
• B. $3\text{cm}$
• C. $4\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is B $3\text{cm}$

Given that,
$p=16\sin \left(\frac{\pi }{3}x\right)\cos \left(64t\right)........\left(1\right)$
General equation of standing wave in an open organ pipe is given by
$p=p_0\sin \left(kx\right)\cos \left(\omega t\right).....\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$ we get,
$k=\frac{2\pi }{\lambda }=\frac{\pi }{3}$
$\Rightarrow \lambda =6\text{cm}$

Pressure node is obtained at: $x=\frac{n\lambda }{2}\text{where}n=1,2,3,\mathrm{4......}$

So, first pressure node is obtained when $n=1$,
i.e $x=\frac{\lambda }{2}=\frac{6}{2}=3\text{cm}$