Question

A standing wave in a pipe of length $L = 1.2m$ is described by $y\left( {x,t} \right) = {y_0}\sin \left[ {\left( {\dfrac{{2\pi }}{L}} \right)x} \right]\sin \left[ {2\left( {\dfrac{\pi }{L}} \right)x + \dfrac{\pi }{4}} \right]$ , based on above information , which one of the following statement is incorrect.(speed of sound in air is $300m{s^{ - 1}}$ )
(A) The pipe is closed at both the ends.
(B) The wavelength of the wave could be $1.2m$ .
(C) There could be a node at $x = 0$ and the antinode at $x = \dfrac{L}{2}$ .
(D) The frequency of the fundamental mode of vibrations is $137.5Hz$ .

Answer 1

Hint : In this question, we are going to first look at the equation given for the standing wave and then operate on the equation by comparing it with the general equation and find the wavelength and find the position of the nodes of that standing wave. Then, the options are analyzed to find incorrect.
The general equation for the standing wave is
$y = {A_0}\sin kx\cos \omega t$
$k = \dfrac{{2\pi }}{\lambda }$ , where $k$ is the propagation wave vector.

Complete Step By Step Answer:
As we are given in the question, with a pipe of length $L = 1.2m$ and the wave equation for it is:
$y\left( {x,t} \right) = {y_0}\sin \left[ {\left( {\dfrac{{2\pi }}{L}} \right)x} \right]\sin \left[ {2\left( {\dfrac{\pi }{L}} \right)x + \dfrac{\pi }{4}} \right]$
The general equation for the standing wave is
$y = {A_0}\sin kx\cos \omega t$
So, for the given equation, if compared with the general equation, we get
kx = \dfrac{{2\pi x}}{L} \\\ \Rightarrow k = \dfrac{{2\pi }}{L} \\\
Where $k$ is the propagation wave vector.
Also we know that, $k = \dfrac{{2\pi }}{\lambda }$
Comparing this with the above obtained value, we get the value of the wavelength as $\lambda = L$ .
Therefore, $\lambda = 1.2m$
The wave equation is given by
$y\left( {x,t} \right) = {y_0}\sin \left[ {\left( {\dfrac{{2\pi }}{L}} \right)x} \right]\sin \left[ {2\left( {\dfrac{\pi }{L}} \right)x + \dfrac{\pi }{4}} \right]$
At $x = 0$ ,
y\left( {0,t} \right) = {y_0}\sin \left[ {\left( {\dfrac{{2\pi }}{L}} \right) \times 0} \right]\sin \left[ {2\left( {\dfrac{\pi }{L}} \right) \times 0 + \dfrac{\pi }{4}} \right] \\\ \Rightarrow y\left( {0,t} \right) = {y_0} \times 0 \times \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow y\left( {0,t} \right) = 0 \\\
At $x = \dfrac{L}{2}$ , if we find the displacement,
$y\left( {\dfrac{L}{2},t} \right) = {y_0}\sin \left[ {\left( {\dfrac{{2\pi }}{L}} \right) \times \dfrac{L}{2}} \right]\sin \left[ {2\left( {\dfrac{\pi }{L}} \right) \times \dfrac{L}{2} + \dfrac{\pi }{4}} \right] \\\ \Rightarrow y\left( {\dfrac{L}{2},t} \right) = {y_0} \times \sin \pi \times \sin \dfrac{{5\pi }}{4} \\\ \Rightarrow y\left( {\dfrac{L}{2},t} \right) = {y_0} \times 0 \times \sin \dfrac{{5\pi }}{4} \\\ \Rightarrow y\left( {\dfrac{L}{2},t} \right) = 0 \\\$
Similarly for $x = L$ ,
$y\left( {L,t} \right) = 0$
Thus, we can say that at $x = 0,\dfrac{L}{2},L$ , we have nodes for the standing wave.
The frequency, for the wave is calculated as:
$\nu = \dfrac{v}{\lambda }$
Where $v$ is the speed of the sound given as $300m{s^{ - 1}}$
$\lambda = 1.2m$ , thus,
$\nu = \dfrac{{300}}{{1.2}} = 250Hz$
So, looking at the four options, we see that the only incorrect option is (D) The frequency of the fundamental mode of vibrations is $137.5Hz$ because the calculated frequency is $250Hz$ .

Note :
Standing wave is a wave that oscillates in time but whose peak amplitude profile does not move in space. The peak amplitude of the wave oscillations at any point in space is constant with time, and the oscillations at different points throughout the wave are in phase. The standing waves are produced when both the ends of the pipe are closed.