Question

A closed organ pipe of length $1.2\text{m}$ vibrates in its first overtone mode. The pressure variation is maximum at :
[neglect end correction]

• A. $0.4\text{m}$ from the open end
• B. $0.4\text{m}$ from the closed end
• C. Both (a) and (b)
• D. $0.8\text{m}$ from the open end

Answer 1

The correct option is C Both (a) and (b)

Modes of vibration of a closed organ pipe are given by
$f_n=\frac{nv}{4l}$ where $n=1,3,\mathrm{5...}....\left(1\right)$

In first overtone mode, pressure variation in the air is shown below.


By substituting, $n=3$ and $v=f\lambda$ in equation $\left(1\right)$, we get,

$\lambda =1.6\text{m}$

Maximum pressure variation from open end,

$l_1=\frac{\lambda }{4}=\frac{1.6}{4}=0.4\text{m}$

Maximum pressure variation from closed end,

$l_1=\frac{\lambda }{2}=\frac{1.6}{2}=0.8\text{m}$

Hence, option $\left(A\right)$ is the correct answer.