Question

A pipe closed at one end has length $83cm$. The number of possible natural oscillations of air column whose frequencies lie below $100Hz$ are (velocity of sound in air $= 332m{s^{ - 1}}$)
(A) $0$
(B) $4$
(C) $5$
(D) $6$

Answer 1

Here, you are given that the pipe is closed. It is a closed organ pipe, basically a cylindrical tube having an air column and one end is closed. Consider sound waves being sent in the closed organ pipe and observe the pressure wave and then derive an equation for natural frequencies. Here, you are given the length of the organ pipe, figure out a way to find the number of natural oscillations possible.

Complete step by step answer:
Consider a closed pipe in which sound waves are being sent. The pressure waves first hit the closed end of the pipe and get reflected and travel back to the open end. When the wave hits the closed end, the same equal and exact pressure is given by the wall of the closed end by Newton’s third law.

As it is receiving what it has given, the reflection is completed without any change of phase. Now this reflected wave travels back to the open end and is reflected again. But this time the end is open and hence it does not get what it has actually given, therefore, for this reflection, there is a change of phase by $180^\circ$. This wave travels back to the closed end and then interferes with the new wave sent by the source.

Now, the phase difference between the old wave and the new wave is $\left( {2l} \right)\left( {\dfrac{{2\pi }}{\lambda }} \right) + \pi$, first term is the phase difference due to distance travelled and second is due to reflection on open end. For constructive interference, $\dfrac{{4\pi l}}{\lambda } + \pi = 2n\pi \to l = \left( {2n - 1} \right)\dfrac{\lambda }{4}$, where $n = 1,2,3,....$. It can also be written as $l = \left( {2n + 1} \right)\dfrac{\lambda }{4}$, where $n = 0,1,2,.....$.

So, for standing wave, frequency will be given by
f = \dfrac{v}{\lambda } \\\ \Rightarrow f = \left( {2n + 1} \right)\dfrac{v}{l} \\\
Let us come back to our question. The frequency in the closed pipe will be $f = \left( {2\left( 0 \right) + 1} \right)\dfrac{{332}}{{4 \times 0.83}} = 100Hz$. As you can see that the minimum value of frequency is supposed to be $100Hz$, so, frequency less than $100Hz$ is not possible. Therefore, the number of possible natural oscillations of air columns whose frequencies lie below $100Hz$ are $0$.

Hence, option A is correct.

Note: First thing you should always do is to convert the unit of the quantities given to you appropriately according to the given situation. Here, length of pipe was given is centimetre but the formula you derived for frequency works in centimetres only when the speed is also given in centimetres per second. You should keep in mind that the normal frequencies for a closed pipe is given as $f = \left( {2n + 1} \right)\dfrac{v}{{4l}}$ and for open pipe, it is given as $f = n\dfrac{v}{{2l}}$. Because in a closed pipe, if you observe the loops, a half loop will always be present whose length is equal to $\dfrac{\lambda }{4}$.