Question

The vibration of air in an open organ pipe of length $30\text{cm}$ is represented by $\Delta p=8\sin \left(\frac{2\pi x}{25}\right)\cos \left(48\pi t\right),$ where $x,p\text{and}t$ are in $\text{cm},{\text{N/m}}^2$ and $\text{sec}$ respectively. The excess pressure (pressure over atmospheric pressure) amplitude at $x=10\text{cm}$ is
[ Take $\sin \left(\frac{4\pi }{5}\right)=0.587$ ]

• A. $5{\text{N/m}}^2$
• B. $4{\text{N/m}}^2$
• C. $4.5{\text{N/m}}^2$
• D. $4.7{\text{N/m}}^2$

Answer 1

The correct option is D $4.7{\text{N/m}}^2$

Given,
$\Delta p=8\sin \left(\frac{2\pi x}{25}\right)\cos \left(48\pi t\right).....\left(1\right)$

Comparing the above equation with,
$\Delta p=2\Delta p_0\sin \left(kx\right)\cos \left(\omega t\right).......\left(2\right)$

Excess pressure amplitude at distance $x$,
$\Delta p\left(x\right)=8\sin \left(\frac{2\pi x}{25}\right)$

So, at $x=10\text{cm}$,
$\Delta p\left(x\right)=8\sin (\frac{2\pi }{25}\times 10)=8\sin \left(4\pi /5\right)=4.7{\text{N/m}}^2$

Thus, option (d) is the correct answer.