Question

A tube open at both ends has fundamental frequency ‘$n$’. If one end of the tube is dipped in liquid to ${\left( {\dfrac{1}{4}} \right)^{th}}$of its length, its new fundamental frequency ${n_1}$ is
A. ${n_1} = \dfrac{n}{2}$
B. ${n_1} = n$
C. ${n_1} = 2n$
D. ${n_1} = 4n$

Answer 1

Hint: In the first case the tube will act as an open pipe, having only two antinodes. The length of the tube divided by 2 will give us the wavelength of the sound traveling. Then when the tube is dipped in water the respective wavelength will be obtained dividing its length by how much portion is dipped in water. The two equations formed when equated in the equation for the speed of sound will give us the required answer.
Formula Used:
Length of the rod, $L = \dfrac{\lambda }{x}$
where $x$ is the number of antinodes formed,
and $\lambda $ is the wavelength of the sound.
Speed of sound, ${\text{v}} = n\lambda $
where $n$ is the fundamental frequency of the tube.

Complete step by step answer:
Standing waves are a combination of two stationary waves such that these two waves are moving in the opposite direction having sample frequency and amplitude and when superimposed on each other their energies are either canceled out or added together.
When a tube is open at both ends having length L, then the standing wave produced in it has two antinodes also known as pressure nodes at both ends. This is known as the first harmonic or the fundamental harmonic.
The tube of length L is initially open at both ends, in this case, let sound’s wavelength be${\lambda _1}$
$\eqalign{
  & L = \dfrac{{{\lambda _1}}}{2} \cr
  & \Rightarrow {\lambda _1} = 2L \cdots \cdots \cdots \left( 1 \right) \cr} $
When this tube gets dipped in water, such that ${\left( {\dfrac{1}{4}} \right)^{th}}$of its length is immersed in it.
So, the tube now acts as a closed pipe, with sound travelling with a wavelength${\lambda _2}$
$\eqalign{
  & \dfrac{L}{4} = \dfrac{{{\lambda _2}}}{4} \cr
  & \Rightarrow {\lambda _2} = 4 \times \dfrac{L}{4} \cr
  & \Rightarrow {\lambda _2} = L \cdots \cdots \cdots \left( 2 \right) \cr} $
Given:
 The fundamental frequency in case I is $n$
The fundamental frequency in case II is ${n_1}$
Now, the speed of sound in both the cases remains the same.
i.e., the speed of sound when the tube is open at both ends and when it is dipped in water such that it acts as a closed pipe remains the same.
$\eqalign{
  & {\text{Speed of sound, v}} = n{\lambda _1} = {n_1}{\lambda _2} \cr
  & \Rightarrow {n_1} = \dfrac{{n{\lambda _1}}}{{{\lambda _2}}} \cr} $
Substituting values of both wavelength from equation (1) and (2), we get:
$\eqalign{
  & {n_1} = \dfrac{{n2L}}{L} \cr
  & \therefore {n_1} = 2n \cr} $
Therefore, the correct option is C i.e., the new fundamental frequency of the tube is${n_1} = 2n$.

Note: There are no missing harmonics in an open organ pipe. The quality of sound from an open organ pipe is, therefore, richer than that from a closed organ pipe in which all the even harmonics of the fundamental are missing.