Question

In an organ pipe whose one end is at $x=0$ , the pressure is expressed by $\text{p} = \text{p}_{0} cos \frac{3 \pi \text{x}}{2} sin 3 0 0 \pi \text{t}$ where x is in meter and $t$ in seconds. The organ pipe can be

• A. Closed at one end, open at another with length $=0.5m$
• B. Open at both ends, length $=1m$
• C. Closed at both ends, length $=2m$
• D. Closed at end, open at another with length $=\frac{2}{3}\text{m}$

Answer 1

Answer: (C)

For $\text{x} = \lambda $
$\frac{3 \pi \text{x}}{2} = 2 \pi $
$⇒ \, \, \, \lambda = \frac{4}{3}$
at x = 0
there is a pressure antinode. So first end must be close and in case of close at bot end
$l = \frac{\text{n} \lambda }{2} \Rightarrow l ⁡ = \frac{2 \text{n}}{3}$
for close at one end
$l = \left(2 \text{n} + 1\right) \frac{\lambda }{4} = \frac{\left(2 \text{n} + 1\right)}{3}$
So for close at one end possible lengths are
$\frac{\left(2 \text{n} + 1\right)}{3} = \frac{1}{3} \text{, } 1 \text{, } \frac{5}{3} \text{, } \frac{7}{3}$
Open at both ends is not possible for close at both end possible length are
$\frac{2 \text{n}}{3} = \frac{2}{3} \text{, } \frac{4}{3} \text{, } 2 \text{, } \frac{8}{3}$ ...... so only possibility is (3)