Question

The vibration of air column in an open organ pipe of length $120\text{cm}$ is represented by $\Delta p=8\sin \left(\frac{2\pi x}{15}\right)\cos \left(48\pi t\right)$ where $x_1,\Delta p$ and $t$ are in $\text{cm}$, ${\text{N/m}}^2$ and $\text{seconds}$ respectively. Then:

• A. The excess pressure amplitude at $x=2.5\text{cm}$ is $4\sqrt{3}{\text{N/m}}^2$.
• B. The excess pressure amplitude at $x=2.5\text{cm}$ is $8\sqrt{3}{\text{N/m}}^2$.
• C. Pressure nodes are located at $7.5\text{cm},15\text{cm},22.5\text{cm}$ from the end.
• D. Pressure nodes are located at $5\text{cm},10\text{cm},15\text{cm}$ from the end.

Answer 1

Answer: (A), (C)

Pressure nodes are located at $7.5\text{cm},15\text{cm},22.5\text{cm}$ from the end.

Given that,
$\Delta p=8\sin \left(\frac{2\pi x}{15}\right)\cos \left(48\pi t\right).........\left(1\right)$
The equation of standing wave in an open organ pipe is given by
$\Delta p=2\Delta p_m\sin \left(kx\right)\cos \left(\omega t\right).........\left(2\right)$

Comparing equation $\left(1\right)$ and $\left(2\right)$,

Excess pressure $\Delta p\left(x\right)=8\sin \left(\frac{2\pi x}{15}\right)$

At $x=\frac{5}{2}\text{m}$,

$\Delta p\left(x\right)=8\sin (\frac{2\pi }{15}\times \frac{5}{2})$

$\Rightarrow \Delta p\left(x\right)=8\sin \left(\frac{\pi }{3}\right)=8\times \frac{\sqrt{3}}{2}=4\sqrt{3}{\text{N/m}}^2$

and $k=\frac{2\pi }{\lambda }=\frac{2\pi }{15}\Rightarrow \frac{\lambda }{2}=7.5\text{cm}$

Pressure nodes will be located at $x=\frac{n\lambda }{2}\text{where}n=0,1,2,\mathrm{3....}$

So, pressure nodes are
For $n=0,x=0\text{cm}$
$n=1,x=7.5\text{cm}$
$n=2,x=15\text{cm}$
$n=3,x=22.5\text{cm}$

Hence, options (a) and (c) are the correct answers.